💡To determine the length of segment AB in the three-dimensional Cartesian coordinate system, we use the distance formula. The distance between two points ( A(x_1, y_1, z_1) ) and ( B(x_2, y_2, z_2) ) is given by:[|AB| = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}]Given points ( A(1, 0, 1) ) and ( B(-1, 1, 2) ), we substitute their coordinates into the formula:[|AB| = sqrt{(-1 - 1)^2 + (1 - 0)^2 + (2 - 1)^2}]Calculating each term:[(-1 - 1)^2 = (-2)^2 = 4][(1 - 0)^2 = 1^2 = 1][(2 - 1)^2 = 1^2 = 1]Adding these results:[4 + 1 + 1 = 6]Taking the square root:[|AB| = sqrt{6}]

💡Okay, so I have this function here: f(x) = cos(x)(2√3 sin(x) - cos(x)) - (1/2)cos(2x) + 1/2. I need to find the smallest positive period and the intervals where the function is monotonically increasing. Then, for part two, I have to find the sum of all real roots of the equation f(x) = a, where a is between -1 and 0, within the interval [0, 2π].First, let me try to simplify the function f(x). It looks a bit complicated, but maybe I can use some trigonometric identities to make it simpler.Starting with f(x):f(x) = cos(x)(2√3 sin(x) - cos(x)) - (1/2)cos(2x) + 1/2Let me expand the first term:cos(x) * 2√3 sin(x) = 2√3 sin(x) cos(x)cos(x) * (-cos(x)) = -cos²(x)So, f(x) becomes:2√3 sin(x) cos(x) - cos²(x) - (1/2)cos(2x) + 1/2Hmm, I remember that sin(2x) = 2 sin(x) cos(x), so 2√3 sin(x) cos(x) is √3 sin(2x). Let me write that:√3 sin(2x) - cos²(x) - (1/2)cos(2x) + 1/2Now, I also recall that cos²(x) can be written using the double angle identity: cos²(x) = (1 + cos(2x))/2. Let me substitute that in:√3 sin(2x) - (1 + cos(2x))/2 - (1/2)cos(2x) + 1/2Simplify the terms:First term: √3 sin(2x)Second term: - (1 + cos(2x))/2Third term: - (1/2)cos(2x)Fourth term: + 1/2Let me combine the constants and the cos(2x) terms:- (1 + cos(2x))/2 - (1/2)cos(2x) + 1/2Let me distribute the negative sign in the second term:-1/2 - (cos(2x))/2 - (cos(2x))/2 + 1/2Combine the constants: -1/2 + 1/2 = 0Combine the cos(2x) terms: - (cos(2x))/2 - (cos(2x))/2 = -cos(2x)So, the entire expression simplifies to:√3 sin(2x) - cos(2x)So, f(x) = √3 sin(2x) - cos(2x)Hmm, that looks like a single sinusoidal function. Maybe I can write it in the form A sin(2x + φ) or A cos(2x + φ). Let me try that.The general form is A sin(θ) + B cos(θ) = C sin(θ + φ), where C = √(A² + B²) and tan(φ) = B/A.Wait, in this case, it's √3 sin(2x) - cos(2x). So, A = √3, B = -1.So, the amplitude C = √( (√3)² + (-1)² ) = √(3 + 1) = √4 = 2.Then, tan(φ) = B/A = (-1)/√3 = -1/√3. So, φ is the angle whose tangent is -1/√3. That's -30 degrees or -π/6 radians.Therefore, f(x) can be written as:2 sin(2x - π/6)Because sin(θ - π/6) = sinθ cos(π/6) - cosθ sin(π/6), which would give the coefficients √3 and -1 when multiplied by 2.So, f(x) = 2 sin(2x - π/6)Alright, that's much simpler.Now, for part (1), the smallest positive period of f(x). Since f(x) is a sine function with argument 2x - π/6, the period of sin(kx + c) is 2π / |k|. Here, k is 2, so the period is 2π / 2 = π.So, the smallest positive period is π.Next, the interval of monotonic increase. Since f(x) is a sine function, it's periodic and its increasing and decreasing intervals repeat every period. Let me find where the function is increasing.The derivative f’(x) will tell me where the function is increasing or decreasing. Let me compute f’(x):f(x) = 2 sin(2x - π/6)f’(x) = 2 * cos(2x - π/6) * 2 = 4 cos(2x - π/6)Wait, no. Wait, derivative of sin(u) is cos(u) * u', so:f’(x) = 2 * cos(2x - π/6) * 2 = 4 cos(2x - π/6)Wait, that seems too big. Wait, f(x) = 2 sin(2x - π/6), so f’(x) = 2 * cos(2x - π/6) * 2 = 4 cos(2x - π/6). Yeah, that's correct.So, f’(x) = 4 cos(2x - π/6). The function is increasing when f’(x) > 0, which is when cos(2x - π/6) > 0.So, cos(θ) > 0 when θ is in (-π/2 + 2πk, π/2 + 2πk) for integers k.So, set 2x - π/6 in that interval:-π/2 + 2πk < 2x - π/6 < π/2 + 2πkLet me solve for x:Add π/6 to all parts:-π/2 + π/6 + 2πk < 2x < π/2 + π/6 + 2πkSimplify the constants:-π/2 + π/6 = (-3π/6 + π/6) = (-2π/6) = -π/3π/2 + π/6 = (3π/6 + π/6) = 4π/6 = 2π/3So:-π/3 + 2πk < 2x < 2π/3 + 2πkDivide all parts by 2:-π/6 + πk < x < π/3 + πkSo, the function f(x) is increasing on intervals (-π/6 + πk, π/3 + πk) for integers k.Since the function is periodic with period π, these intervals repeat every π units.So, in terms of intervals, the function is increasing from -π/6 + πk to π/3 + πk.But since we are dealing with x ∈ ℝ, the function is increasing on each interval of length (π/3 - (-π/6)) = π/3 + π/6 = π/2, and then decreasing for the next π/2, and so on.But the question asks for the interval of monotonic increase. So, the function is increasing on each interval (-π/6 + πk, π/3 + πk) for any integer k.So, that's part (1). The smallest positive period is π, and the function is increasing on intervals (-π/6 + πk, π/3 + πk) for integers k.Now, moving on to part (2). We need to find the sum of all real roots of the equation f(x) = a, where -1 < a < 0, within the interval [0, 2π].Given that f(x) = 2 sin(2x - π/6), and a is between -1 and 0, so f(x) = a implies 2 sin(2x - π/6) = a, so sin(2x - π/6) = a/2.Since a is between -1 and 0, a/2 is between -0.5 and 0. So, sin(θ) = c, where c is between -0.5 and 0.We need to find all x in [0, 2π] such that sin(2x - π/6) = a/2.Let me denote θ = 2x - π/6. Then, θ ranges from 2*0 - π/6 = -π/6 to 2*2π - π/6 = 4π - π/6 = (24π/6 - π/6) = 23π/6.So, θ ∈ [-π/6, 23π/6].We need to solve sin(θ) = a/2, where a/2 ∈ (-0.5, 0). So, sin(θ) = c, c ∈ (-0.5, 0).The general solution for sin(θ) = c is θ = arcsin(c) + 2πk or θ = π - arcsin(c) + 2πk for integers k.But since c is negative, arcsin(c) will be negative. Let me denote α = arcsin(c), so α ∈ (-π/6, 0) because c ∈ (-0.5, 0). Because sin(-π/6) = -0.5.So, θ = α + 2πk or θ = π - α + 2πk.But θ must be in [-π/6, 23π/6]. Let me find all solutions θ in this interval.First, let's find all possible k such that θ = α + 2πk is in [-π/6, 23π/6].Similarly for θ = π - α + 2πk.Given that α is between -π/6 and 0, let's see:For θ = α + 2πk:We need α + 2πk ≥ -π/6 and α + 2πk ≤ 23π/6.Since α ≥ -π/6, the smallest k can be is k=0: θ = α, which is ≥ -π/6.For the upper bound: α + 2πk ≤ 23π/6.Since α ≤ 0, 2πk ≤ 23π/6 - α ≤ 23π/6.So, 2πk ≤ 23π/6 ⇒ k ≤ 23/12 ≈ 1.916. So, k can be 0, 1.Similarly, for k=0: θ = α ∈ [-π/6, 0]For k=1: θ = α + 2π ∈ [2π - π/6, 2π] = [11π/6, 2π]But 2π is 12π/6, so 11π/6 to 12π/6.But our θ goes up to 23π/6, which is more than 2π (which is 12π/6). Wait, 23π/6 is 3π + 5π/6, which is more than 2π. So, actually, θ can go beyond 2π.Wait, but x is in [0, 2π], so θ = 2x - π/6 is in [-π/6, 4π - π/6] = [-π/6, 23π/6]. So, θ can go up to 23π/6, which is more than 2π.So, for θ = α + 2πk, k can be 0, 1, 2, because 2π*2 = 4π, which is less than 23π/6 (≈3.833π). Wait, 23π/6 is approximately 3.833π, so 4π is larger. So, k can be 0,1,2.Wait, let's check:For k=2: θ = α + 4π. Since α ≥ -π/6, θ = α + 4π ≥ 4π - π/6 = 23π/6, which is exactly the upper limit. So, θ can be up to 23π/6.So, k=0: θ ∈ [-π/6, 0]k=1: θ ∈ [2π - π/6, 2π] = [11π/6, 12π/6]k=2: θ ∈ [4π - π/6, 4π] = [23π/6, 24π/6]But our θ only goes up to 23π/6, so for k=2, θ = α + 4π must be ≤23π/6.Since α ≤0, θ = α +4π ≤4π.But 4π =24π/6, which is greater than 23π/6.So, θ = α +4π must be ≤23π/6 ⇒ α ≤23π/6 -4π=23π/6 -24π/6= -π/6.But α ≥ -π/6, so α must be exactly -π/6.But since a ∈ (-1,0), c =a/2 ∈ (-0.5,0). So, sin(θ)=c, c ∈ (-0.5,0). So, arcsin(c) ∈ (-π/6,0).Therefore, for k=2, θ = α +4π. Since α > -π/6, θ = α +4π >4π - π/6=23π/6.But our θ can only go up to 23π/6, so θ=23π/6 is the maximum. So, θ=α +4π can only reach 23π/6 when α=-π/6, but since α > -π/6, θ=α +4π >23π/6, which is outside our interval. Therefore, for k=2, θ would be beyond 23π/6, which is not allowed. So, k=2 doesn't give a solution within our interval.Similarly, for k=1: θ = α +2π. Since α ∈ (-π/6,0), θ ∈ (2π - π/6, 2π) = (11π/6, 12π/6). But 12π/6=2π, which is the upper limit of θ.Wait, but θ can go up to 23π/6, which is more than 2π. So, actually, θ=α +2πk for k=0,1,2.But for k=2, θ=α +4π. Since α > -π/6, θ=α +4π >4π - π/6=23π/6, which is beyond our θ's upper limit. So, only k=0 and k=1 give solutions within θ ∈ [-π/6,23π/6].Similarly, for θ=π - α +2πk.Again, let's find k such that θ=π - α +2πk is within [-π/6,23π/6].Since α ∈ (-π/6,0), π - α ∈ (π, π + π/6)= (π,7π/6).So, θ=π - α +2πk.We need θ ∈ [-π/6,23π/6].So, for k=0: θ=π - α ∈ (π,7π/6) which is within [-π/6,23π/6].For k=1: θ=π - α +2π=3π - α ∈ (3π -0,3π - (-π/6))=(3π, 19π/6). Since 19π/6≈3.166π, which is less than 23π/6≈3.833π.For k=2: θ=π - α +4π=5π - α ∈ (5π,5π +π/6)= (5π,31π/6). But 31π/6≈5.166π, which is greater than 23π/6≈3.833π. So, θ=5π - α would be beyond our interval.Therefore, k=0 and k=1 give solutions within θ ∈ [-π/6,23π/6].So, in total, for each k=0 and k=1, we have two solutions from θ=α +2πk and θ=π - α +2πk.But wait, for each k, we have two solutions? Or for each k, one solution from each case.Wait, no. For each k, θ=α +2πk and θ=π - α +2πk are two different solutions.So, for k=0:θ1=αθ2=π - αFor k=1:θ3=α +2πθ4=π - α +2πSo, in total, four solutions for θ in [-π/6,23π/6].Therefore, four solutions for θ, which correspond to four solutions for x in [0,2π].So, let's write down the four solutions:1. θ1=α ⇒ 2x - π/6=α ⇒ x=(α + π/6)/22. θ2=π - α ⇒ 2x - π/6=π - α ⇒ 2x=π - α + π/6=7π/6 - α ⇒ x=(7π/6 - α)/23. θ3=α +2π ⇒ 2x - π/6=α +2π ⇒ 2x=α +2π + π/6=α +13π/6 ⇒ x=(α +13π/6)/24. θ4=π - α +2π=3π - α ⇒ 2x - π/6=3π - α ⇒ 2x=3π - α + π/6=19π/6 - α ⇒ x=(19π/6 - α)/2So, we have four solutions:x1=(α + π/6)/2x2=(7π/6 - α)/2x3=(α +13π/6)/2x4=(19π/6 - α)/2Now, we need to find the sum of these x's.Let me compute x1 +x2 +x3 +x4:x1 +x2 +x3 +x4 = [(α + π/6)/2] + [(7π/6 - α)/2] + [(α +13π/6)/2] + [(19π/6 - α)/2]Let me combine the terms:= [α + π/6 +7π/6 - α + α +13π/6 +19π/6 - α]/2Simplify numerator:α - α + α - α + π/6 +7π/6 +13π/6 +19π/6The α terms cancel out: (α - α + α - α)=0Now, the constants:π/6 +7π/6 +13π/6 +19π/6 = (1 +7 +13 +19)π/6 = (40)π/6 = 20π/3So, x1 +x2 +x3 +x4 = (20π/3)/2 =10π/3Wait, but 20π/3 divided by 2 is 10π/3.But wait, 20π/3 is approximately 6.666π, divided by 2 is 3.333π, which is 10π/3.But wait, 10π/3 is approximately 3.333π, but our interval is [0,2π], which is about 6.283.Wait, but the sum of the roots is 10π/3, which is approximately 10.472, but 2π is approximately 6.283, so 10π/3 is about 1.666*2π, which is more than 2π. But the roots are within [0,2π], so their sum can be more than 2π.Wait, but let me double-check my calculation.x1 +x2 +x3 +x4 = [(α + π/6) + (7π/6 - α) + (α +13π/6) + (19π/6 - α)] /2Let me compute numerator:(α + π/6 +7π/6 - α + α +13π/6 +19π/6 - α)Group α terms:α - α + α - α =0Constant terms:π/6 +7π/6 +13π/6 +19π/6 = (1 +7 +13 +19)π/6 =40π/6=20π/3So, numerator is 20π/3, divided by 2 is 10π/3.So, the sum is 10π/3.But wait, 10π/3 is approximately 10.472, but 2π is approximately 6.283, so 10π/3 is about 1.666*2π, which is more than 2π. But the roots are within [0,2π], so their sum can be more than 2π.But let me think again. Since the function f(x)=2 sin(2x - π/6) has period π, in the interval [0,2π], it completes two full periods. Therefore, the equation f(x)=a will have two solutions in each period, so four solutions in total.But when I sum the roots, I get 10π/3, which is approximately 10.472. But 2π is approximately 6.283, so 10π/3 is about 1.666*2π, which is more than 2π.Wait, but in reality, the sum of the roots should be something like 2*(sum of roots in one period). Since in each period, the sum of roots is something, and since there are two periods, the total sum would be twice that.Wait, let me think about the function f(x)=2 sin(2x - π/6). In each period of π, the function goes up and down once. So, in each period, the equation f(x)=a will have two solutions, symmetric around the midpoint of the period.Wait, but actually, for a sinusoidal function, the sum of the roots in one period can be found by considering the function's symmetry.Wait, for f(x)=A sin(Bx + C), the roots of f(x)=a in one period are symmetric around the midpoint of the period.Wait, but in this case, the function is f(x)=2 sin(2x - π/6). So, the period is π, and in each period, the function goes from 0 to π, and then π to 2π, but since the period is π, it's actually from x=0 to x=π, and then x=π to x=2π.Wait, but the function is 2 sin(2x - π/6). So, the phase shift is π/12, because 2x - π/6=0 ⇒ x=π/12.So, the function starts at x=π/12, and then completes a full cycle by x=π/12 + π=13π/12.Wait, but in the interval [0,2π], the function will have two full periods: from π/12 to 13π/12, and from 13π/12 to 25π/12, but 25π/12 is more than 2π (which is 24π/12). So, actually, the second period is from π/12 to 13π/12, and the third period would start at 13π/12, but we only go up to 2π=24π/12.Wait, maybe I'm overcomplicating.Alternatively, since the function has period π, in [0,2π], it completes two periods. So, for each period, the equation f(x)=a will have two solutions, so four solutions in total.Now, for each period, the sum of the roots can be found by considering the function's symmetry.In a sinusoidal function, the sum of the roots in one period for f(x)=a is equal to twice the midpoint of the period.Wait, for example, in the first period, from x=0 to x=π, the midpoint is π/2. So, the sum of the two roots in this period would be 2*(π/2)=π.Similarly, in the second period, from x=π to x=2π, the midpoint is 3π/2, so the sum of the two roots would be 2*(3π/2)=3π.Therefore, the total sum of all four roots would be π +3π=4π.Wait, but earlier I calculated 10π/3≈3.333π, which is less than 4π≈12.566.Hmm, there's a discrepancy here. Let me check my earlier calculation.Wait, in the earlier calculation, I considered θ=2x - π/6, and found four solutions for θ, which correspond to four x's. Then, I summed them up and got 10π/3.But according to the symmetry argument, the sum should be 4π.Wait, maybe my earlier approach was wrong.Wait, let me think again. The function f(x)=2 sin(2x - π/6) has period π. So, in [0,2π], it has two periods.In each period, the equation f(x)=a will have two solutions, symmetric around the midpoint of the period.So, for the first period, from x=0 to x=π, the midpoint is π/2. So, the two roots are symmetric around π/2, so their sum is 2*(π/2)=π.Similarly, for the second period, from x=π to x=2π, the midpoint is 3π/2, so the two roots sum to 2*(3π/2)=3π.Therefore, the total sum is π +3π=4π.But in my earlier calculation, I got 10π/3≈3.333π, which is less than 4π.So, which one is correct?Wait, let me check my earlier calculation again.I had:x1=(α + π/6)/2x2=(7π/6 - α)/2x3=(α +13π/6)/2x4=(19π/6 - α)/2Sum: [ (α + π/6) + (7π/6 - α) + (α +13π/6) + (19π/6 - α) ] /2Simplify numerator:α - α + α - α + π/6 +7π/6 +13π/6 +19π/6=0 + (1 +7 +13 +19)π/6=40π/6=20π/3So, sum=20π/3 /2=10π/3≈3.333πBut according to the symmetry argument, it should be 4π≈12.566.Wait, but 10π/3≈10.472, which is less than 4π≈12.566.Wait, but 10π/3 is approximately 10.472, which is more than 3π≈9.424.Wait, but 4π≈12.566 is more than 10π/3≈10.472.Wait, so which one is correct?Wait, maybe my symmetry argument is wrong.Wait, let me think about the function f(x)=2 sin(2x - π/6). Let me plot it mentally.The function has amplitude 2, period π, phase shift π/12.So, it starts at x=π/12, goes up to maximum at x=π/12 + π/4= (π/12 + 3π/12)=4π/12=π/3, then back to zero at x=π/12 + π/2=7π/12, then to minimum at x=π/12 + 3π/4=10π/12=5π/6, then back to zero at x=π/12 + π=13π/12.So, in the first period, from π/12 to13π/12, the function completes one full cycle.Similarly, the second period is from13π/12 to25π/12, but since we're only going up to2π=24π/12, the second period is cut off at24π/12.Wait, so in [0,2π], the function completes two full periods: from π/12 to13π/12, and from13π/12 to25π/12, but since25π/12>24π/12=2π, the second period is only partially in [0,2π].Wait, no, because2π=24π/12, so the second period would be from13π/12 to25π/12, but25π/12>24π/12, so the second period is from13π/12 to24π/12, which is less than a full period.Wait, so actually, in [0,2π], the function completes one full period from π/12 to13π/12, and then a partial period from13π/12 to24π/12.But wait, no, because the function is periodic with period π, so from0 to π, it's one period, and from π to2π, it's another period.Wait, but the phase shift complicates things.Wait, maybe it's better to consider that in [0,2π], the function f(x)=2 sin(2x - π/6) has two full periods, each of length π.So, in each period, the equation f(x)=a will have two solutions, so four solutions in total.Now, for each period, the sum of the roots can be found by considering the function's symmetry.In a sinusoidal function, the sum of the roots in one period for f(x)=a is equal to twice the midpoint of the period.Wait, for example, in the first period, from x=0 to x=π, the midpoint is π/2. So, the two roots are symmetric around π/2, so their sum is 2*(π/2)=π.Similarly, in the second period, from x=π to x=2π, the midpoint is 3π/2, so the two roots sum to 2*(3π/2)=3π.Therefore, the total sum of all four roots would be π +3π=4π.But according to my earlier calculation, the sum was10π/3≈3.333π, which is less than4π.Wait, so which one is correct?Wait, maybe my earlier approach was wrong because I didn't account for the phase shift correctly.Wait, let me try another approach.Since f(x)=2 sin(2x - π/6), and we're solving f(x)=a, where a ∈ (-1,0).So, 2 sin(2x - π/6)=a ⇒ sin(2x - π/6)=a/2.Let me denote θ=2x - π/6, so θ=2x - π/6.Then, x=(θ + π/6)/2.We need to find all θ in [2*0 - π/6,2*2π - π/6]=[-π/6,23π/6] such that sin(θ)=a/2.Since a ∈ (-1,0), a/2 ∈ (-0.5,0).So, sin(θ)=c, where c ∈ (-0.5,0).The general solution is θ=arcsin(c)+2πk and θ=π - arcsin(c)+2πk, for integers k.But since c is negative, arcsin(c)= -arcsin(|c|).Let me denote β=arcsin(|c|), so β ∈ (0,π/6), because |c| ∈ (0,0.5), and arcsin(0.5)=π/6.So, arcsin(c)= -β.Therefore, the solutions are:θ= -β +2πk and θ=π +β +2πk.Now, we need θ ∈ [-π/6,23π/6].Let's find all k such that θ= -β +2πk and θ=π +β +2πk are within [-π/6,23π/6].First, for θ= -β +2πk:We need -β +2πk ≥ -π/6 ⇒2πk ≥ -π/6 +β.Since β>0, -π/6 +β > -π/6.So, 2πk ≥ -π/6 +β.Similarly, θ= -β +2πk ≤23π/6 ⇒2πk ≤23π/6 +β.So, k must satisfy:(-π/6 +β)/(2π) ≤k ≤(23π/6 +β)/(2π)Compute:(-π/6 +β)/(2π)= (-1/6 +β/π)/2≈ (-0.1667 +β/π)/2Similarly, (23π/6 +β)/(2π)= (23/6 +β/π)/2≈ (3.833 +β/π)/2≈1.9165 +β/(2π)Since β ∈ (0,π/6), β/π ∈ (0,1/6).So, (-π/6 +β)/(2π)≈(-0.1667 +0)/2≈-0.0833 to (-0.1667 +0.1667)/2≈0.Similarly, (23π/6 +β)/(2π)≈(3.833 +0)/2≈1.9165 to (3.833 +0.1667)/2≈2.0.Therefore, k can be 0,1,2.But let's check:For k=0: θ= -β ∈ (-π/6,0), which is within [-π/6,23π/6].For k=1: θ= -β +2π ∈ (2π - π/6,2π)= (11π/6,12π/6)= (11π/6,2π). But 2π=12π/6, and our θ goes up to23π/6≈3.833π, which is more than2π. So, θ= -β +2π is within [11π/6,2π], which is within [-π/6,23π/6].For k=2: θ= -β +4π ∈ (4π - π/6,4π)= (23π/6,24π/6). But our θ only goes up to23π/6, so θ= -β +4π must be ≤23π/6.But θ= -β +4π=4π -β.Since β>0, 4π -β <4π.But 4π=24π/6, and our θ upper limit is23π/6≈3.833π.So, 4π -β=24π/6 -β.We need 24π/6 -β ≤23π/6 ⇒ -β ≤-π/6 ⇒β ≥π/6.But β=arcsin(|c|), and |c|<0.5, so β<π/6.Therefore, 4π -β >23π/6.So, θ= -β +4π >23π/6, which is outside our interval.Therefore, k=2 does not give a solution within θ ∈ [-π/6,23π/6].Similarly, for θ=π +β +2πk:We need π +β +2πk ∈ [-π/6,23π/6].So, π +β +2πk ≥-π/6 ⇒2πk ≥-π/6 -π -β= -7π/6 -β.And π +β +2πk ≤23π/6 ⇒2πk ≤23π/6 -π -β=17π/6 -β.So, k must satisfy:(-7π/6 -β)/(2π) ≤k ≤(17π/6 -β)/(2π)Compute:(-7π/6 -β)/(2π)= (-7/6 -β/π)/2≈ (-1.1667 -β/π)/2≈-0.5833 -β/(2π)Similarly, (17π/6 -β)/(2π)= (17/6 -β/π)/2≈ (2.833 -β/π)/2≈1.4165 -β/(2π)Since β ∈ (0,π/6), β/π ∈ (0,1/6).So, lower bound≈-0.5833 -0≈-0.5833 to -0.5833 -0.1667≈-0.75Upper bound≈1.4165 -0≈1.4165 to1.4165 -0.1667≈1.25Therefore, k can be 0,1.For k=0: θ=π +β ∈ (π,π +π/6)= (π,7π/6), which is within [-π/6,23π/6].For k=1: θ=π +β +2π=3π +β ∈ (3π,3π +π/6)= (3π,19π/6). Since19π/6≈3.166π, which is less than23π/6≈3.833π.For k=2: θ=π +β +4π=5π +β ∈ (5π,5π +π/6)= (5π,31π/6). But31π/6≈5.166π>23π/6≈3.833π, so θ=5π +β>23π/6, which is outside our interval.Therefore, k=0 and k=1 give solutions within θ ∈ [-π/6,23π/6].So, in total, we have four solutions:From θ= -β +2πk:k=0: θ1= -βk=1: θ2= -β +2πFrom θ=π +β +2πk:k=0: θ3=π +βk=1: θ4=π +β +2π=3π +βSo, four solutions:θ1= -βθ2= -β +2πθ3=π +βθ4=3π +βNow, let's find x for each θ:x=(θ + π/6)/2So,x1=( -β + π/6 )/2x2=( -β +2π + π/6 )/2=( -β +13π/6 )/2x3=(π +β + π/6 )/2=(7π/6 +β)/2x4=(3π +β + π/6 )/2=(19π/6 +β)/2Now, let's compute the sum x1 +x2 +x3 +x4:= [ (-β + π/6 ) + (-β +13π/6 ) + (7π/6 +β ) + (19π/6 +β ) ] /2Simplify numerator:-β -β +β +β + π/6 +13π/6 +7π/6 +19π/6=0 + (1 +13 +7 +19)π/6=40π/6=20π/3So, sum=20π/3 /2=10π/3≈3.333πWait, but earlier I thought it should be4π≈12.566.But according to this calculation, it's10π/3≈3.333π.But wait, 10π/3≈10.472, which is more than3π≈9.424.Wait, but 10π/3 is approximately3.333π, which is less than4π≈12.566.Wait, I'm getting confused with the units.Wait, 10π/3 is approximately10.472, which is more than3π≈9.424, but less than4π≈12.566.Wait, but in any case, according to the calculation, the sum is10π/3.But according to the symmetry argument, it should be4π.Wait, maybe the symmetry argument is incorrect because of the phase shift.Wait, let me think again.In each period, the function f(x)=2 sin(2x - π/6) has a phase shift of π/12.So, the midpoints of the periods are not at π/2 and3π/2, but shifted.Wait, the first period is from π/12 to13π/12, so the midpoint is (π/12 +13π/12)/2=14π/12/2=7π/12.Similarly, the second period is from13π/12 to25π/12, but since we're only going up to24π/12=2π, the midpoint would be (13π/12 +24π/12)/2=37π/24≈1.541π.Wait, but in reality, the second period is only partially in [0,2π].Wait, maybe the symmetry argument doesn't hold because the second period is cut off.Wait, but in reality, in [0,2π], the function completes two full periods: from0 toπ, and fromπ to2π.Wait, but the phase shift isπ/12, so the function starts atπ/12, not at0.Therefore, in [0,2π], the function doesn't start at a peak or trough, but somewhere in between.Therefore, the symmetry argument about the sum of roots in each period being twice the midpoint may not hold because the periods are not aligned with the interval [0,2π].Therefore, my earlier calculation of10π/3 is correct.Wait, but let me check with specific values.Let me take a specific value of a, say a=0.Then, f(x)=0 ⇒2 sin(2x - π/6)=0 ⇒sin(2x - π/6)=0 ⇒2x - π/6=kπ ⇒x= (kπ +π/6)/2.In [0,2π], k can be0,1,2,3,4.Wait, for k=0: x=π/12≈0.2618k=1: x=(π +π/6)/2=7π/12≈1.8326k=2: x=(2π +π/6)/2=13π/12≈3.4002k=3: x=(3π +π/6)/2=19π/12≈5.0000k=4: x=(4π +π/6)/2=25π/12≈6.5449, which is beyond2π≈6.2832.So, in [0,2π], the roots are atπ/12,7π/12,13π/12,19π/12.Sum=π/12 +7π/12 +13π/12 +19π/12= (1+7+13+19)π/12=40π/12=10π/3≈3.333π.Which matches my earlier calculation.But according to the symmetry argument, if I consider each period, the sum should be twice the midpoint.But in this case, the sum is10π/3≈3.333π, which is less than4π.Wait, but when a=0, the roots are at the points where the function crosses zero, which are symmetric around the midpoints of the periods.Wait, but in this case, the midpoints of the periods are atπ/2 and3π/2.But the roots are atπ/12,7π/12,13π/12,19π/12.Sum=10π/3.So, in this specific case, the sum is10π/3.Therefore, my earlier calculation is correct.Therefore, for any a ∈ (-1,0), the sum of the roots in [0,2π] is10π/3.Wait, but when a=0, the sum is10π/3, which is approximately3.333π.But when a approaches-1, the roots approach the minima of the function, which are at5π/6 and19π/6.Wait, but let me check.Wait, the function f(x)=2 sin(2x - π/6) has minima at2x - π/6=3π/2 +2πk ⇒x= (3π/2 +π/6)/2 +πk= (10π/6)/2 +πk=5π/6 +πk.So, in [0,2π], the minima are at5π/6≈2.618 and5π/6 +π=11π/6≈5.759.Similarly, the maxima are at2x - π/6=π/2 +2πk ⇒x=(π/2 +π/6)/2 +πk=(2π/3)/2 +πk=π/3 +πk.So, in [0,2π], maxima atπ/3≈1.047 andπ/3 +π=4π/3≈4.188.So, when a approaches-1, the roots approach5π/6 and11π/6.Similarly, when a approaches0, the roots approach the midpoints between the maxima and minima.But regardless of the value of a ∈ (-1,0), the sum of the roots remains10π/3.Wait, is that possible?Wait, in the case when a=0, the sum is10π/3.When a approaches-1, the roots approach5π/6 and11π/6, but there are four roots, so two approaching5π/6 and11π/6, and the other two approaching somewhere else?Wait, no, when a approaches-1, the equation f(x)=a has two roots near the minima, but since a is approaching-1 from above, the roots would be near5π/6 and11π/6, but there are two more roots near the other side.Wait, no, actually, for a ∈ (-1,0), the equation f(x)=a has four roots in [0,2π]: two near the minima and two near the maxima.Wait, no, actually, for a ∈ (-1,0), the function f(x)=2 sin(2x - π/6) will cross the line y=a four times in [0,2π]: twice on the rising part and twice on the falling part.But regardless, the sum of the roots remains10π/3.Wait, that seems counterintuitive because when a changes, the positions of the roots change, but their sum remains constant.But in the specific case when a=0, the sum is10π/3, and when a approaches-1, the sum remains10π/3.Wait, that must be because the function is symmetric in such a way that the sum of the roots remains constant regardless of a.Wait, let me think about it.In general, for a sinusoidal function f(x)=A sin(Bx + C), the sum of the roots of f(x)=a over one period is equal to twice the midpoint of the period.But in this case, because of the phase shift and the interval [0,2π], the sum over two periods is10π/3.Wait, but in the specific case when a=0, the sum is10π/3, which is consistent with my earlier calculation.Therefore, regardless of the value of a ∈ (-1,0), the sum of the roots in [0,2π] is10π/3.Therefore, the answer is10π/3.Wait, but in my earlier calculation, I got10π/3, but when I thought about the symmetry, I thought it should be4π.But according to the specific case when a=0, the sum is10π/3, which is approximately3.333π, which is less than4π.Therefore, the correct answer is10π/3.So, the sum of all real roots of the equation f(x)=a within [0,2π] is10π/3.

💡Alright, so I have these three questions about counting the number of ways to choose classes for social practice activities. Let me try to figure them out step by step.Starting with question (1): "How many different ways are there to choose students from 1 class to participate in social practice activities?" Hmm, okay. So, the school has three grades: first year with 14 classes, second year with 14 classes, and third year with 15 classes. I need to choose just one class from any of these grades. So, if I think about it, there are three separate groups of classes: first year, second year, and third year. For each group, I can choose one class. So, for the first year, there are 14 classes, so 14 ways. Similarly, for the second year, 14 ways, and for the third year, 15 ways. Since these are separate choices, I should add them up. So, 14 (first year) + 14 (second year) + 15 (third year). Let me calculate that: 14 + 14 is 28, and 28 + 15 is 43. So, there are 43 different ways to choose one class from any grade. That seems straightforward.Moving on to question (2): "How many different ways are there to choose students from one class in each grade to participate in social practice activities?" Okay, so this time, instead of choosing just one class from any grade, I need to choose one class from each grade. That means one class from the first year, one from the second year, and one from the third year.So, for the first year, there are 14 classes, so 14 choices. For the second year, also 14 classes, so 14 choices. For the third year, 15 classes, so 15 choices. Since these are independent choices, I should multiply the number of choices for each grade together. So, 14 (first year) × 14 (second year) × 15 (third year). Let me compute that: 14 × 14 is 196, and 196 × 15. Hmm, 196 × 10 is 1960, and 196 × 5 is 980, so adding those together gives 1960 + 980 = 2940. So, there are 2940 different ways to choose one class from each grade.Now, question (3): "How many different ways are there to choose students from 2 classes to participate in social practice activities, with the requirement that these 2 classes are from different grades?" Okay, so this is a bit more complex. I need to choose two classes, but they have to be from different grades. So, they can't be both from the first year, both from the second year, or both from the third year. They have to be from two different grades.So, let me think about the possible pairs of grades. There are three grades, so the possible pairs are first and second, first and third, and second and third. So, three different cases.First case: choosing one class from the first year and one from the second year. There are 14 classes in the first year and 14 in the second year, so the number of ways is 14 × 14.Second case: choosing one class from the first year and one from the third year. There are 14 classes in the first year and 15 in the third year, so that's 14 × 15.Third case: choosing one class from the second year and one from the third year. There are 14 classes in the second year and 15 in the third year, so that's 14 × 15.Now, since these are separate cases, I should add them up. So, total number of ways is (14 × 14) + (14 × 15) + (14 × 15). Let me compute each part:14 × 14 is 196.14 × 15 is 210.So, adding them up: 196 + 210 + 210. Let me compute that step by step. 196 + 210 is 406, and 406 + 210 is 616.So, there are 616 different ways to choose two classes from different grades.Wait, just to make sure I didn't make a mistake. Let me double-check the calculations.For the first case: 14 × 14 is indeed 196.Second case: 14 × 15 is 210.Third case: 14 × 15 is also 210.Adding them together: 196 + 210 is 406, plus another 210 gives 616. Yep, that seems correct.So, summarizing:1. Choosing one class from any grade: 43 ways.2. Choosing one class from each grade: 2940 ways.3. Choosing two classes from different grades: 616 ways.I think that's all. It makes sense because for the first question, it's just adding up all the classes, for the second, it's multiplying the choices across grades, and for the third, it's considering all possible pairs of grades and adding those products together.**Final Answer**(1) The number of ways is boxed{43}.(2) The number of ways is boxed{2940}.(3) The number of ways is boxed{616}.

💡First, I need to determine how many pencils Shreehari should give to each student so that everyone receives the same amount.He has a total of 125 pencils and there are 25 students in the class.To find out the number of pencils per student, I will divide the total number of pencils by the number of students.So, I will calculate 125 divided by 25.Performing the division, 125 ÷ 25 equals 5.Therefore, Shreehari should give each student 5 pencils.

💡Alright, so I've got this geometry problem here about an equilateral triangle ABC with an incircle of radius 2. The center of this incircle is point I. There's a point P such that the distance from P to I is 1, and I need to find the maximum value of the ratio of the areas of triangles APB and APC. Hmm, okay, let me try to break this down step by step.First, since ABC is an equilateral triangle, all its sides are equal, and all its angles are 60 degrees. The incircle touches each side of the triangle, and the center I is also the centroid, right? Wait, no, in an equilateral triangle, the centroid, circumcenter, inradius, and orthocenter all coincide, so yes, I is the centroid as well. That might be useful later.The radius of the incircle is given as 2. I remember that the radius of the incircle (r) of an equilateral triangle is related to its side length (a) by the formula r = (a√3)/6. So, if r = 2, then 2 = (a√3)/6, which means a = (2*6)/√3 = 12/√3 = 4√3. So, the side length of the triangle is 4√3. That's good to know.Now, point P is such that PI = 1. So, P lies on a circle centered at I with radius 1. Since I is the centroid, and the centroid divides the medians in a 2:1 ratio, the distance from each vertex to I is 2/3 of the median length. Wait, the median length in an equilateral triangle is also the altitude, right? The altitude (h) of an equilateral triangle is h = (a√3)/2. So, h = (4√3 * √3)/2 = (4*3)/2 = 6. So, the distance from each vertex to I is 2/3 of 6, which is 4. So, AI = BI = CI = 4.So, P is a point inside the triangle such that its distance from I is 1. I need to find the maximum ratio of the areas of triangles APB and APC. Hmm, areas of triangles... I remember that the area of a triangle can be given by (1/2)*ab*sinθ, where a and b are sides and θ is the included angle. Maybe that will come into play here.Let me try to visualize this. Since ABC is equilateral, all sides are 4√3, and all angles are 60 degrees. Point I is the centroid, so it's located at the intersection of the medians, each of length 6. Point P is somewhere inside the triangle, 1 unit away from I. I need to find the position of P such that the ratio of the areas of APB to APC is maximized.Maybe I can parameterize the position of P. Let me consider coordinate geometry. Let's place the triangle ABC in a coordinate system to make things easier. Let me set point A at (0, 6), point B at (-2√3, 0), and point C at (2√3, 0). Wait, is that correct? Let me check.If the side length is 4√3, then the base BC should be 4√3 long. If I place point B at (-2√3, 0) and point C at (2√3, 0), then the distance between B and C is 4√3, which is correct. The height of the triangle is 6, so point A is at (0, 6). The centroid I is at the average of the coordinates of A, B, and C, so I_x = (0 + (-2√3) + 2√3)/3 = 0, and I_y = (6 + 0 + 0)/3 = 2. So, I is at (0, 2). That makes sense because the centroid is 1/3 of the height from the base.So, point P is somewhere on the circle centered at (0, 2) with radius 1. I need to find the coordinates of P such that the ratio of the areas of APB to APC is maximized.Let me denote point P as (x, y). Then, the area of triangle APB can be found using the determinant formula:Area of APB = (1/2) | (Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By) ) |Wait, no, that's the area of triangle ABC. For triangle APB, it's similar but with points A, P, B.So, coordinates:A = (0, 6)P = (x, y)B = (-2√3, 0)So, the area of triangle APB is (1/2) | (0*(y - 0) + x*(0 - 6) + (-2√3)*(6 - y) ) | = (1/2) | 0 + x*(-6) + (-2√3)*(6 - y) | = (1/2) | -6x -12√3 + 2√3 y |.Similarly, the area of triangle APC is (1/2) | (0*(y - 0) + x*(0 - 6) + (2√3)*(6 - y) ) | = (1/2) | 0 + x*(-6) + 2√3*(6 - y) | = (1/2) | -6x + 12√3 - 2√3 y |.So, the ratio of areas is:[ (1/2) | -6x -12√3 + 2√3 y | ] / [ (1/2) | -6x + 12√3 - 2√3 y | ] = | -6x -12√3 + 2√3 y | / | -6x + 12√3 - 2√3 y |.Simplify numerator and denominator:Numerator: | -6x -12√3 + 2√3 y | = | -6x + 2√3 y -12√3 |.Denominator: | -6x + 12√3 - 2√3 y | = | -6x - 2√3 y +12√3 |.Hmm, that's a bit messy. Maybe there's a better way to approach this.Alternatively, since all the points are in a plane, maybe I can use vectors or coordinate geometry more effectively.Wait, another thought: in an equilateral triangle, the areas of triangles APB and APC can be related to the distances from P to the sides AB and AC, respectively. Because the area is proportional to the height from P to each side.But since ABC is equilateral, the heights from P to AB and AC would be related to the angles at A. Maybe I can express the areas in terms of the angles.Let me denote θ as the angle between AP and the median AI. Since AI is the median, it's also the altitude and the angle bisector. So, if I can express the areas in terms of θ, maybe I can find the ratio.Wait, but P is constrained to be on the circle of radius 1 around I. So, maybe I can parameterize P in terms of an angle φ from the center I.Let me try that. Let me set up polar coordinates with I at (0, 2). Then, point P can be represented as (r cos φ, r sin φ + 2), where r = 1. So, P = (cos φ, sin φ + 2).Wait, no, in Cartesian coordinates, if I is at (0, 2), then P is at (cos φ, sin φ + 2). Because the circle is centered at (0, 2) with radius 1.So, point P is (cos φ, sin φ + 2). Now, I can plug this into the area expressions.First, let's compute the area of triangle APB.Points:A = (0, 6)P = (cos φ, sin φ + 2)B = (-2√3, 0)Using the shoelace formula:Area APB = (1/2) | (0*(sin φ + 2 - 0) + cos φ*(0 - 6) + (-2√3)*(6 - (sin φ + 2)) ) |= (1/2) | 0 + cos φ*(-6) + (-2√3)*(4 - sin φ) |= (1/2) | -6 cos φ -8√3 + 2√3 sin φ |.Similarly, area APC:Points:A = (0, 6)P = (cos φ, sin φ + 2)C = (2√3, 0)Area APC = (1/2) | 0*(sin φ + 2 - 0) + cos φ*(0 - 6) + 2√3*(6 - (sin φ + 2)) |= (1/2) | 0 + cos φ*(-6) + 2√3*(4 - sin φ) |= (1/2) | -6 cos φ + 8√3 - 2√3 sin φ |.So, the ratio is:[ | -6 cos φ -8√3 + 2√3 sin φ | ] / [ | -6 cos φ + 8√3 - 2√3 sin φ | ].Hmm, that's still a bit complicated, but maybe I can factor out some terms.Let me factor out -6 cos φ from both numerator and denominator:Numerator: | -6 cos φ -8√3 + 2√3 sin φ | = | -6 cos φ + 2√3 sin φ -8√3 |.Denominator: | -6 cos φ + 8√3 - 2√3 sin φ | = | -6 cos φ - 2√3 sin φ +8√3 |.Hmm, not sure if that helps. Maybe I can write this as:Numerator: | -6 cos φ + 2√3 sin φ -8√3 |.Denominator: | -6 cos φ - 2√3 sin φ +8√3 |.Let me denote this as:Numerator: | A sin φ + B cos φ + C |.Wait, actually, it's linear in sin φ and cos φ. Maybe I can write it in the form R sin(φ + α) + C or something like that.Let me consider the numerator:-6 cos φ + 2√3 sin φ -8√3.This can be written as:2√3 sin φ -6 cos φ -8√3.Similarly, the denominator:-6 cos φ -2√3 sin φ +8√3.Which is:-2√3 sin φ -6 cos φ +8√3.So, both numerator and denominator have terms of sin φ and cos φ, plus constants.I can write these as:Numerator: 2√3 sin φ -6 cos φ -8√3.Denominator: -2√3 sin φ -6 cos φ +8√3.Let me factor out the coefficients of sin φ and cos φ:Numerator: 2√3 sin φ -6 cos φ -8√3.Let me factor out 2 from the first two terms:2(√3 sin φ - 3 cos φ) -8√3.Similarly, denominator:-2√3 sin φ -6 cos φ +8√3.Factor out -2 from the first two terms:-2(√3 sin φ + 3 cos φ) +8√3.Hmm, maybe I can write √3 sin φ ± 3 cos φ as a single sine or cosine function.Recall that a sin φ + b cos φ = R sin(φ + α), where R = √(a² + b²) and tan α = b/a.So, for the numerator:√3 sin φ - 3 cos φ.Here, a = √3, b = -3.So, R = √( (√3)^2 + (-3)^2 ) = √(3 + 9) = √12 = 2√3.And tan α = b/a = (-3)/√3 = -√3. So, α = -60 degrees or 300 degrees.So, √3 sin φ - 3 cos φ = 2√3 sin(φ - 60°).Similarly, for the denominator:√3 sin φ + 3 cos φ.Here, a = √3, b = 3.So, R = √(3 + 9) = √12 = 2√3.tan α = b/a = 3/√3 = √3. So, α = 60 degrees.Thus, √3 sin φ + 3 cos φ = 2√3 sin(φ + 60°).So, substituting back into numerator and denominator:Numerator: 2*(2√3 sin(φ - 60°)) -8√3 = 4√3 sin(φ - 60°) -8√3.Denominator: -2*(2√3 sin(φ + 60°)) +8√3 = -4√3 sin(φ + 60°) +8√3.So, the ratio becomes:|4√3 sin(φ - 60°) -8√3| / | -4√3 sin(φ + 60°) +8√3 |.Factor out 4√3 from numerator and denominator:Numerator: 4√3 | sin(φ - 60°) - 2 |.Denominator: 4√3 | -sin(φ + 60°) + 2 |.So, the 4√3 cancels out:| sin(φ - 60°) - 2 | / | -sin(φ + 60°) + 2 |.Simplify the expressions inside the absolute values:Numerator: | sin(φ - 60°) - 2 |. Since sin(φ - 60°) ranges between -1 and 1, so sin(φ - 60°) - 2 ranges between -3 and -1. So, the absolute value is 2 - sin(φ - 60°).Denominator: | -sin(φ + 60°) + 2 | = | 2 - sin(φ + 60°) |. Similarly, sin(φ + 60°) ranges between -1 and 1, so 2 - sin(φ + 60°) ranges between 1 and 3. So, the absolute value is just 2 - sin(φ + 60°).So, the ratio simplifies to:(2 - sin(φ - 60°)) / (2 - sin(φ + 60°)).Now, let's denote θ = φ - 60°, so φ = θ + 60°. Then, φ + 60° = θ + 120°.So, the ratio becomes:(2 - sin θ) / (2 - sin(θ + 120°)).Now, let's compute sin(θ + 120°). Using the sine addition formula:sin(θ + 120°) = sin θ cos 120° + cos θ sin 120°.We know that cos 120° = -1/2 and sin 120° = √3/2.So,sin(θ + 120°) = sin θ*(-1/2) + cos θ*(√3/2) = (-1/2) sin θ + (√3/2) cos θ.Therefore, the ratio becomes:(2 - sin θ) / (2 - [ (-1/2) sin θ + (√3/2) cos θ ]) = (2 - sin θ) / (2 + (1/2) sin θ - (√3/2) cos θ).Let me multiply numerator and denominator by 2 to eliminate the fractions:Numerator: 2*(2 - sin θ) = 4 - 2 sin θ.Denominator: 2*(2 + (1/2) sin θ - (√3/2) cos θ) = 4 + sin θ - √3 cos θ.So, the ratio is now:(4 - 2 sin θ) / (4 + sin θ - √3 cos θ).Hmm, this seems a bit more manageable. Let me denote this ratio as R:R = (4 - 2 sin θ) / (4 + sin θ - √3 cos θ).I need to find the maximum value of R as θ varies. To maximize R, I can consider it as a function of θ and find its maximum.One approach is to take the derivative of R with respect to θ and set it to zero, but that might be a bit involved. Alternatively, I can express the denominator in terms of a single sine or cosine function, which might make it easier to find the maximum.Let me focus on the denominator: 4 + sin θ - √3 cos θ.This can be written as 4 + (sin θ - √3 cos θ).Again, similar to earlier, I can write sin θ - √3 cos θ as a single sine function.Let me denote:sin θ - √3 cos θ = R sin(θ - α),where R = √(1^2 + (√3)^2) = √(1 + 3) = 2,and tan α = (√3)/1 = √3, so α = 60°.Therefore,sin θ - √3 cos θ = 2 sin(θ - 60°).So, the denominator becomes:4 + 2 sin(θ - 60°).Similarly, the numerator is 4 - 2 sin θ.So, R = (4 - 2 sin θ) / (4 + 2 sin(θ - 60°)).Simplify numerator and denominator by factoring out 2:Numerator: 2*(2 - sin θ).Denominator: 2*(2 + sin(θ - 60°)).Cancel out the 2s:R = (2 - sin θ) / (2 + sin(θ - 60°)).Now, let me express sin θ in terms of sin(θ - 60° + 60°). Using the sine addition formula:sin θ = sin(θ - 60° + 60°) = sin(θ - 60°) cos 60° + cos(θ - 60°) sin 60°.We know that cos 60° = 1/2 and sin 60° = √3/2.So,sin θ = sin(θ - 60°)*(1/2) + cos(θ - 60°)*(√3/2).Let me denote φ = θ - 60°, so θ = φ + 60°.Then,sin θ = sin φ*(1/2) + cos φ*(√3/2).So, substituting back into R:R = (2 - [ (1/2) sin φ + (√3/2) cos φ ]) / (2 + sin φ).Let me write this as:R = [2 - (1/2 sin φ + √3/2 cos φ)] / (2 + sin φ).Let me denote the numerator as N and the denominator as D:N = 2 - (1/2 sin φ + √3/2 cos φ).D = 2 + sin φ.So,R = N / D.I need to find the maximum of R. Let's express N in terms of D.Wait, maybe I can write N as:N = 2 - (1/2 sin φ + √3/2 cos φ).Let me factor out 1/2:N = 2 - (1/2)(sin φ + √3 cos φ).Again, sin φ + √3 cos φ can be written as 2 sin(φ + 60°), since:sin φ + √3 cos φ = 2 sin(φ + 60°).Because,R = √(1^2 + (√3)^2) = 2,and tan α = √3/1 = √3, so α = 60°.Therefore,sin φ + √3 cos φ = 2 sin(φ + 60°).So,N = 2 - (1/2)*2 sin(φ + 60°) = 2 - sin(φ + 60°).Thus, R becomes:R = (2 - sin(φ + 60°)) / (2 + sin φ).Hmm, interesting. So, R is expressed in terms of φ, which is θ - 60°, which itself is related to the original angle φ we used earlier. Wait, this might be getting a bit convoluted. Let me see if I can find another approach.Alternatively, maybe I can set t = sin φ, and express R in terms of t, then find the maximum.But before that, let me think about the range of φ. Since φ = θ - 60°, and θ can vary such that P lies on the circle around I, which is inside the triangle ABC. So, φ can vary over some range, but I'm not sure exactly what it is. Maybe it's from -60° to 60°, but I need to confirm.Wait, actually, since P is inside the triangle, and I is the centroid, P can't be too far from I. But since PI = 1, and the distance from I to any side is 2 (since the inradius is 2), P is inside the incircle, so it's within 1 unit from I. So, φ can vary such that P remains inside the triangle.But perhaps instead of worrying about the range of φ, I can consider R as a function and find its maximum.So, R = (2 - sin(φ + 60°)) / (2 + sin φ).Let me denote ψ = φ + 60°, so φ = ψ - 60°.Then,R = (2 - sin ψ) / (2 + sin(ψ - 60°)).Again, using the sine subtraction formula:sin(ψ - 60°) = sin ψ cos 60° - cos ψ sin 60° = (1/2) sin ψ - (√3/2) cos ψ.So,R = (2 - sin ψ) / [2 + (1/2 sin ψ - √3/2 cos ψ)].Simplify denominator:2 + (1/2 sin ψ - √3/2 cos ψ) = 2 + (1/2 sin ψ) - (√3/2 cos ψ).So,R = (2 - sin ψ) / [2 + (1/2 sin ψ) - (√3/2 cos ψ)].Again, this seems similar to before. Maybe I can express the denominator in terms of sin ψ and cos ψ.Alternatively, perhaps I can set t = sin ψ, and express cos ψ in terms of t, but that might complicate things.Wait, another idea: since R is a function of ψ, maybe I can write it as:R = (2 - sin ψ) / [2 + (1/2 sin ψ - √3/2 cos ψ)].Let me denote A = 1/2 and B = -√3/2, so the denominator becomes 2 + A sin ψ + B cos ψ.So,R = (2 - sin ψ) / (2 + A sin ψ + B cos ψ).To find the maximum of R, I can use calculus. Let me treat R as a function of ψ and find its derivative.Let me denote numerator N = 2 - sin ψ,denominator D = 2 + (1/2) sin ψ - (√3/2) cos ψ.Then,dR/dψ = (N’ D - N D’) / D².Compute N’:N’ = -cos ψ.Compute D’:D’ = (1/2) cos ψ + (√3/2) sin ψ.So,dR/dψ = [ (-cos ψ)(2 + (1/2) sin ψ - (√3/2) cos ψ) - (2 - sin ψ)( (1/2) cos ψ + (√3/2) sin ψ ) ] / D².This looks quite messy, but let's try to compute it step by step.First, compute the first term:(-cos ψ)(2 + (1/2) sin ψ - (√3/2) cos ψ) =-2 cos ψ - (1/2) sin ψ cos ψ + (√3/2) cos² ψ.Second, compute the second term:(2 - sin ψ)( (1/2) cos ψ + (√3/2) sin ψ ) =2*(1/2 cos ψ + √3/2 sin ψ) - sin ψ*(1/2 cos ψ + √3/2 sin ψ) =cos ψ + √3 sin ψ - (1/2 sin ψ cos ψ + √3/2 sin² ψ).So, putting it all together:dR/dψ = [ (-2 cos ψ - (1/2) sin ψ cos ψ + (√3/2) cos² ψ ) - (cos ψ + √3 sin ψ - (1/2 sin ψ cos ψ + √3/2 sin² ψ )) ] / D².Simplify the numerator:First, distribute the negative sign:-2 cos ψ - (1/2) sin ψ cos ψ + (√3/2) cos² ψ - cos ψ - √3 sin ψ + (1/2 sin ψ cos ψ + √3/2 sin² ψ ).Now, combine like terms:-2 cos ψ - cos ψ = -3 cos ψ.- (1/2) sin ψ cos ψ + (1/2 sin ψ cos ψ ) = 0.(√3/2) cos² ψ + √3/2 sin² ψ = (√3/2)(cos² ψ + sin² ψ) = √3/2.- √3 sin ψ.So, overall, the numerator simplifies to:-3 cos ψ - √3 sin ψ + √3/2.Therefore,dR/dψ = [ -3 cos ψ - √3 sin ψ + √3/2 ] / D².To find critical points, set the numerator equal to zero:-3 cos ψ - √3 sin ψ + √3/2 = 0.Let me rearrange:-3 cos ψ - √3 sin ψ = -√3/2.Multiply both sides by -1:3 cos ψ + √3 sin ψ = √3/2.Let me write this as:3 cos ψ + √3 sin ψ = √3/2.Let me divide both sides by √3 to simplify:(3/√3) cos ψ + sin ψ = 1/2.Simplify 3/√3 = √3.So,√3 cos ψ + sin ψ = 1/2.This is of the form A cos ψ + B sin ψ = C.Where A = √3, B = 1, C = 1/2.We can write this as R cos(ψ - α) = C, where R = √(A² + B²) = √(3 + 1) = 2,and tan α = B/A = 1/√3, so α = 30°.So,2 cos(ψ - 30°) = 1/2.Thus,cos(ψ - 30°) = 1/4.Therefore,ψ - 30° = ± arccos(1/4) + 360°k, where k is integer.So,ψ = 30° ± arccos(1/4) + 360°k.Since ψ is an angle, we can consider the principal values:ψ = 30° + arccos(1/4) and ψ = 30° - arccos(1/4).Compute arccos(1/4):arccos(1/4) ≈ 75.522°.So,ψ ≈ 30° + 75.522° ≈ 105.522°,andψ ≈ 30° - 75.522° ≈ -45.522°.Since angles are periodic, -45.522° is equivalent to 360° - 45.522° ≈ 314.478°.So, the critical points are at ψ ≈ 105.522° and ψ ≈ 314.478°.Now, we need to determine which of these gives a maximum for R.Let me compute R at these critical points.First, compute R at ψ ≈ 105.522°.But before that, let me note that R is a continuous function, and we need to check whether these critical points correspond to maxima or minima.Alternatively, since we're looking for the maximum ratio, perhaps we can evaluate R at these points and see which one gives a higher value.But let me compute R at ψ = 105.522°.First, compute sin ψ:sin(105.522°) ≈ sin(90° + 15.522°) ≈ cos(15.522°) ≈ 0.9613.Wait, actually, sin(105.522°) = sin(90° + 15.522°) = cos(15.522°) ≈ 0.9613.Wait, no, that's not correct. sin(90° + x) = cos x, but sin(105.522°) is sin(90° + 15.522°) = cos(15.522°) ≈ 0.9613.Similarly, sin(ψ) ≈ 0.9613.Compute R = (2 - sin ψ) / (2 + sin(ψ - 60°)).Wait, earlier we had R = (2 - sin ψ) / (2 + sin(ψ - 60°)).Wait, no, actually, earlier substitution led to R = (2 - sin ψ) / (2 + sin(ψ - 60°)).But let me double-check.Wait, earlier, we had:R = (2 - sin ψ) / (2 + sin(ψ - 60°)).But actually, from earlier steps, when we set ψ = φ + 60°, and then expressed R in terms of ψ, we had:R = (2 - sin ψ) / (2 + sin(ψ - 60°)).Yes, that's correct.So, for ψ ≈ 105.522°, compute sin(ψ - 60°) = sin(45.522°) ≈ 0.7133.So,R ≈ (2 - 0.9613) / (2 + 0.7133) ≈ (1.0387) / (2.7133) ≈ 0.382.Similarly, for ψ ≈ 314.478°, compute sin ψ ≈ sin(314.478°) ≈ sin(360° - 45.522°) ≈ -sin(45.522°) ≈ -0.7133.Then, sin(ψ - 60°) = sin(314.478° - 60°) = sin(254.478°) ≈ sin(180° + 74.478°) ≈ -sin(74.478°) ≈ -0.9613.So,R ≈ (2 - (-0.7133)) / (2 + (-0.9613)) ≈ (2 + 0.7133) / (2 - 0.9613) ≈ 2.7133 / 1.0387 ≈ 2.61.So, R is approximately 2.61 at ψ ≈ 314.478°, which is higher than the other critical point. Therefore, the maximum ratio occurs at ψ ≈ 314.478°, giving R ≈ 2.61.But let's see if we can find an exact expression for this maximum.Recall that at the critical point, we had:√3 cos ψ + sin ψ = 1/2.Let me denote this as equation (1):√3 cos ψ + sin ψ = 1/2.We can write this as:2 cos(ψ - 30°) = 1/2,since √3 cos ψ + sin ψ = 2 cos(ψ - 30°).Wait, let me verify:cos(ψ - 30°) = cos ψ cos 30° + sin ψ sin 30° = (√3/2) cos ψ + (1/2) sin ψ.So,2 cos(ψ - 30°) = √3 cos ψ + sin ψ.Therefore, equation (1) becomes:2 cos(ψ - 30°) = 1/2,so,cos(ψ - 30°) = 1/4.Thus,ψ - 30° = ± arccos(1/4) + 360°k,which is what we had earlier.So, the solutions are ψ = 30° ± arccos(1/4) + 360°k.Now, let's compute R at ψ = 30° - arccos(1/4).So, ψ = 30° - arccos(1/4).Compute sin ψ:sin(30° - arccos(1/4)).Using the sine subtraction formula:sin(A - B) = sin A cos B - cos A sin B.Let A = 30°, B = arccos(1/4).So,sin(30° - arccos(1/4)) = sin 30° cos(arccos(1/4)) - cos 30° sin(arccos(1/4)).We know that cos(arccos(1/4)) = 1/4,and sin(arccos(1/4)) = √(1 - (1/4)^2) = √(15/16) = √15/4.Also, sin 30° = 1/2,cos 30° = √3/2.So,sin(30° - arccos(1/4)) = (1/2)(1/4) - (√3/2)(√15/4) = 1/8 - (√45)/8 = 1/8 - (3√5)/8 = (1 - 3√5)/8.Similarly, compute sin(ψ - 60°):ψ - 60° = (30° - arccos(1/4)) - 60° = -30° - arccos(1/4).So,sin(ψ - 60°) = sin(-30° - arccos(1/4)) = -sin(30° + arccos(1/4)).Using sine addition formula:sin(30° + arccos(1/4)) = sin 30° cos(arccos(1/4)) + cos 30° sin(arccos(1/4)).= (1/2)(1/4) + (√3/2)(√15/4) = 1/8 + (√45)/8 = 1/8 + (3√5)/8 = (1 + 3√5)/8.Therefore,sin(ψ - 60°) = - (1 + 3√5)/8.So, now, compute R:R = (2 - sin ψ) / (2 + sin(ψ - 60°)).Substitute the values:sin ψ = (1 - 3√5)/8,sin(ψ - 60°) = - (1 + 3√5)/8.So,Numerator: 2 - (1 - 3√5)/8 = (16/8 - 1/8 + 3√5/8) = (15 + 3√5)/8.Denominator: 2 + (- (1 + 3√5)/8 ) = (16/8 - 1/8 - 3√5/8 ) = (15 - 3√5)/8.Thus,R = (15 + 3√5)/8 divided by (15 - 3√5)/8 = (15 + 3√5)/(15 - 3√5).Factor out 3 from numerator and denominator:= 3(5 + √5) / 3(5 - √5) = (5 + √5)/(5 - √5).Multiply numerator and denominator by (5 + √5):= (5 + √5)^2 / (25 - 5) = (25 + 10√5 + 5) / 20 = (30 + 10√5)/20 = (3 + √5)/2.So, R = (3 + √5)/2 ≈ (3 + 2.236)/2 ≈ 5.236/2 ≈ 2.618, which matches our earlier approximate calculation.Therefore, the maximum value of the ratio R is (3 + √5)/2.So, the maximum ratio of the areas of triangles APB to APC is (3 + √5)/2.

💡Okay, so I have this trigonometry problem here, and I need to find the value of this expression: (sin2x + 2sin²x)/(1 - tanx), given that cos(x + π/4) = 3/5 and x is between 17π/12 and 7π/4. Hmm, that seems a bit complicated, but let me try to break it down step by step.First, I remember that cos(x + π/4) can be expanded using the cosine addition formula. The formula is cos(A + B) = cosA cosB - sinA sinB. So, applying that here, cos(x + π/4) should be cosx * cos(π/4) - sinx * sin(π/4). I know that cos(π/4) and sin(π/4) are both √2/2, so that simplifies things a bit.Let me write that out:cos(x + π/4) = cosx * (√2/2) - sinx * (√2/2) = (√2/2)(cosx - sinx).And we're told that this equals 3/5. So,(√2/2)(cosx - sinx) = 3/5.To solve for cosx - sinx, I can multiply both sides by 2/√2:cosx - sinx = (3/5) * (2/√2) = (6)/(5√2).Hmm, maybe I should rationalize the denominator here. Multiplying numerator and denominator by √2 gives:cosx - sinx = (6√2)/(5*2) = (3√2)/5.Okay, so cosx - sinx equals 3√2/5. Got that.Now, I think squaring both sides might help because I can use the Pythagorean identity. Let's try that:(cosx - sinx)² = (3√2/5)².Expanding the left side:cos²x - 2sinx cosx + sin²x = (9*2)/25 = 18/25.But cos²x + sin²x is 1, so substituting that in:1 - 2sinx cosx = 18/25.Subtracting 18/25 from both sides:1 - 18/25 = 2sinx cosx.Calculating 1 - 18/25: 25/25 - 18/25 = 7/25.So, 2sinx cosx = 7/25.Wait, 2sinx cosx is sin2x, right? So, sin2x = 7/25.Alright, that's useful. I can note that down.Now, going back to the original expression I need to find: (sin2x + 2sin²x)/(1 - tanx). Let me write that out:(sin2x + 2sin²x)/(1 - tanx).I know sin2x is 7/25, so that's helpful. But I still need to find sin²x and tanx.Hmm, maybe I can find sinx and cosx individually. I have cosx - sinx = 3√2/5 and I also know that cos²x + sin²x = 1. Maybe I can set up a system of equations.Let me denote cosx as C and sinx as S for simplicity.So, I have:1. C - S = 3√2/52. C² + S² = 1I can solve this system to find C and S.From equation 1, I can express C as:C = S + 3√2/5.Substituting this into equation 2:(S + 3√2/5)² + S² = 1.Expanding the square:S² + 2*S*(3√2/5) + (3√2/5)² + S² = 1.Simplify:2S² + (6√2/5)S + (9*2)/25 = 1.Calculating (9*2)/25: 18/25.So, 2S² + (6√2/5)S + 18/25 = 1.Subtract 1 from both sides:2S² + (6√2/5)S + 18/25 - 1 = 0.Calculating 18/25 - 1: 18/25 - 25/25 = -7/25.So, the equation becomes:2S² + (6√2/5)S - 7/25 = 0.This is a quadratic in terms of S. Let me write it as:2S² + (6√2/5)S - 7/25 = 0.To make it easier, multiply all terms by 25 to eliminate denominators:50S² + 30√2 S - 7 = 0.Now, using the quadratic formula to solve for S:S = [-30√2 ± √( (30√2)^2 - 4*50*(-7) )]/(2*50).Calculating discriminant:(30√2)^2 = 900*2 = 1800.4*50*7 = 1400.So, discriminant is 1800 + 1400 = 3200.√3200 = √(64*50) = 8√50 = 8*5√2 = 40√2.So, S = [-30√2 ± 40√2]/100.Calculating both possibilities:1. S = (-30√2 + 40√2)/100 = (10√2)/100 = √2/10.2. S = (-30√2 - 40√2)/100 = (-70√2)/100 = -7√2/10.So, sinx could be √2/10 or -7√2/10.Now, I need to figure out which one is correct based on the given domain for x: 17π/12 < x < 7π/4.Let me convert these to degrees to get a better sense.17π/12 is equal to (17*15) = 255 degrees.7π/4 is equal to (7*45) = 315 degrees.So, x is between 255 degrees and 315 degrees. That places x in the fourth quadrant, where sine is negative and cosine is positive.Therefore, sinx should be negative. So, sinx = -7√2/10.Alright, so sinx = -7√2/10.Now, let's find cosx. From equation 1:C = S + 3√2/5.So, C = (-7√2/10) + 3√2/5.Convert 3√2/5 to tenths: 6√2/10.So, C = (-7√2 + 6√2)/10 = (-√2)/10.Wait, that can't be right. If x is in the fourth quadrant, cosx should be positive, but this gives cosx = -√2/10, which is negative. That contradicts.Hmm, did I make a mistake somewhere?Let me check my calculations.From equation 1: C = S + 3√2/5.If S = -7√2/10, then C = (-7√2/10) + 3√2/5.Convert 3√2/5 to tenths: 6√2/10.So, C = (-7√2 + 6√2)/10 = (-√2)/10.Hmm, that's negative. But in the fourth quadrant, cosine should be positive. So, perhaps I made a mistake in choosing the sign when solving the quadratic.Wait, when I solved for S, I got two solutions: √2/10 and -7√2/10. I chose -7√2/10 because sine is negative in the fourth quadrant. But if that leads to cosine being negative, which is not possible, then maybe I need to reconsider.Wait, perhaps I messed up the quadratic solution.Let me double-check the quadratic equation.We had:2S² + (6√2/5)S - 7/25 = 0.Multiplying by 25:50S² + 30√2 S - 7 = 0.Quadratic formula: S = [-b ± √(b² - 4ac)]/(2a).Here, a = 50, b = 30√2, c = -7.Discriminant: b² - 4ac = (30√2)^2 - 4*50*(-7) = 1800 + 1400 = 3200.√3200 = 40√2.So, S = [-30√2 ± 40√2]/100.So,1. S = (-30√2 + 40√2)/100 = (10√2)/100 = √2/10.2. S = (-30√2 - 40√2)/100 = (-70√2)/100 = -7√2/10.So, that's correct.But in the fourth quadrant, sine is negative, so S = -7√2/10.Then, C = S + 3√2/5 = (-7√2/10) + 6√2/10 = (-√2)/10.But cosine is negative here, which contradicts the fourth quadrant where cosine is positive.Hmm, so maybe I made a mistake in the earlier step.Wait, let's go back.We had cos(x + π/4) = 3/5.Given x is between 17π/12 and 7π/4, which is between 255 degrees and 315 degrees.So, x + π/4 is between 17π/12 + π/4 = 17π/12 + 3π/12 = 20π/12 = 5π/3, and 7π/4 + π/4 = 8π/4 = 2π.So, x + π/4 is between 5π/3 and 2π, which is between 300 degrees and 360 degrees, so in the fourth quadrant.In the fourth quadrant, cosine is positive, which matches cos(x + π/4) = 3/5.But when we found cosx - sinx = 3√2/5, and then tried to solve for cosx and sinx, we ended up with cosx negative, which is conflicting.Wait, perhaps I need to consider the sign when taking square roots.Earlier, when I squared cosx - sinx = 3√2/5, I got (cosx - sinx)^2 = 18/25, which led to 1 - 2sinx cosx = 18/25, so 2sinx cosx = 7/25.But then, when I tried to find cosx + sinx, I think I might have missed something.Wait, let me think. I have cosx - sinx = 3√2/5, and I also have 2sinx cosx = 7/25.If I can find cosx + sinx, then I can solve for both cosx and sinx.Because (cosx + sinx)^2 = cos²x + 2sinx cosx + sin²x = 1 + 2sinx cosx.We know 2sinx cosx = 7/25, so (cosx + sinx)^2 = 1 + 7/25 = 32/25.Therefore, cosx + sinx = ±√(32/25) = ±(4√2)/5.Now, since x is in the fourth quadrant, cosx is positive and sinx is negative. So, cosx + sinx could be positive or negative depending on the magnitudes.But let's think about x + π/4 being in the fourth quadrant, between 5π/3 and 2π.So, x + π/4 is in the fourth quadrant, which is between 300 degrees and 360 degrees.In that case, sin(x + π/4) would be negative because sine is negative in the fourth quadrant.But sin(x + π/4) = sinx cos(π/4) + cosx sin(π/4) = (sinx + cosx)(√2/2).So, sin(x + π/4) = (sinx + cosx)(√2/2).Since sin(x + π/4) is negative, (sinx + cosx) must be negative.Therefore, cosx + sinx = -4√2/5.Alright, so now I have:cosx - sinx = 3√2/5,cosx + sinx = -4√2/5.Now, I can solve these two equations for cosx and sinx.Adding the two equations:(cosx - sinx) + (cosx + sinx) = 3√2/5 + (-4√2/5),2cosx = (-√2)/5,cosx = (-√2)/10.Wait, but earlier I thought cosx should be positive in the fourth quadrant. Hmm, this is conflicting.Wait, but if cosx = (-√2)/10, that would mean cosx is negative, which is not possible in the fourth quadrant.Hmm, something is wrong here.Wait, let's double-check.We have x in the fourth quadrant, so cosx is positive and sinx is negative.But from the equations:cosx - sinx = 3√2/5,cosx + sinx = -4√2/5.Adding them gives 2cosx = (-√2)/5 => cosx = (-√2)/10, which is negative.But that contradicts the fourth quadrant where cosx should be positive.So, perhaps I made a mistake in determining the sign of cosx + sinx.Wait, earlier I said that since x + π/4 is in the fourth quadrant, sin(x + π/4) is negative, so (sinx + cosx)(√2/2) is negative, which implies sinx + cosx is negative.But if sinx + cosx is negative, and cosx is positive, sinx must be negative enough to make the sum negative.But in our case, cosx is negative, which would make sinx + cosx even more negative, but we have cosx positive in the fourth quadrant.Wait, I'm getting confused.Let me try another approach.Given that x is in the fourth quadrant, so cosx is positive, sinx is negative.We have:cosx - sinx = 3√2/5,andcosx + sinx = -4√2/5.But if cosx is positive and sinx is negative, then cosx + sinx could be positive or negative depending on the magnitudes.But according to the earlier reasoning, sin(x + π/4) is negative, so (sinx + cosx) must be negative.Therefore, cosx + sinx = -4√2/5.So, even though cosx is positive, sinx is negative enough to make the sum negative.So, let's proceed with that.So, from the two equations:1. cosx - sinx = 3√2/5,2. cosx + sinx = -4√2/5.Let me solve these two equations.Adding them:2cosx = (3√2/5 - 4√2/5) = (-√2)/5,So, cosx = (-√2)/10.But wait, cosx is negative here, which contradicts the fourth quadrant where cosx should be positive.Hmm, this is a problem.Wait, maybe I made a mistake in the sign when solving for cosx + sinx.Let me think again.We have:(cosx + sinx)^2 = 32/25,so cosx + sinx = ±4√2/5.But since x + π/4 is in the fourth quadrant, sin(x + π/4) is negative, which implies (sinx + cosx) is negative.Therefore, cosx + sinx = -4√2/5.But if cosx is positive and sinx is negative, can their sum be negative?Yes, if the magnitude of sinx is greater than cosx.So, let's proceed.So, cosx = (-√2)/10, which is negative, but that contradicts the fourth quadrant.Wait, maybe I made a mistake in the earlier step.Wait, let me solve the two equations again.Equation 1: cosx - sinx = 3√2/5,Equation 2: cosx + sinx = -4√2/5.Let me add them:2cosx = (3√2/5 - 4√2/5) = (-√2)/5,So, cosx = (-√2)/10.But cosx is negative, which is not possible in the fourth quadrant.Hmm, so perhaps I made a mistake in determining the sign of cosx + sinx.Wait, maybe cosx + sinx is positive?But earlier reasoning suggested it's negative because sin(x + π/4) is negative.Wait, let me double-check.sin(x + π/4) = sinx cos(π/4) + cosx sin(π/4) = (sinx + cosx)(√2/2).Given that x + π/4 is in the fourth quadrant, sin(x + π/4) is negative, so (sinx + cosx)(√2/2) is negative, which implies sinx + cosx is negative.Therefore, cosx + sinx must be negative.But if cosx is positive and sinx is negative, their sum can be negative if |sinx| > cosx.But in our case, solving the equations gives cosx negative, which is conflicting.Wait, perhaps I made a mistake in the quadratic solution.Let me go back to the quadratic equation.We had:50S² + 30√2 S - 7 = 0.Using quadratic formula:S = [-30√2 ± √( (30√2)^2 - 4*50*(-7) )]/(2*50).Calculating discriminant:(30√2)^2 = 900*2 = 1800,4*50*7 = 1400,So, discriminant = 1800 + 1400 = 3200,√3200 = 40√2.Thus,S = [-30√2 ± 40√2]/100.So,1. S = (-30√2 + 40√2)/100 = (10√2)/100 = √2/10,2. S = (-30√2 - 40√2)/100 = (-70√2)/100 = -7√2/10.So, S = √2/10 or S = -7√2/10.But in the fourth quadrant, sinx is negative, so S = -7√2/10.Then, from equation 1: C = S + 3√2/5 = (-7√2/10) + 6√2/10 = (-√2)/10.So, cosx = -√2/10.But cosx is negative, which contradicts the fourth quadrant.Hmm, this is confusing.Wait, maybe the initial assumption that cosx - sinx = 3√2/5 is correct, but when solving for cosx and sinx, we end up with cosx negative, which is conflicting.Is there another way to approach this problem without solving for cosx and sinx individually?Let me think.We have to find (sin2x + 2sin²x)/(1 - tanx).We already found sin2x = 7/25.So, let's write the expression as:(sin2x + 2sin²x)/(1 - tanx) = (7/25 + 2sin²x)/(1 - tanx).But I still need to find sin²x and tanx.Alternatively, maybe we can express everything in terms of sinx and cosx.Let me try that.So,Numerator: sin2x + 2sin²x = 2sinx cosx + 2sin²x = 2sinx (cosx + sinx).Denominator: 1 - tanx = 1 - (sinx/cosx) = (cosx - sinx)/cosx.So, the expression becomes:[2sinx (cosx + sinx)] / [(cosx - sinx)/cosx] = [2sinx (cosx + sinx) * cosx] / (cosx - sinx).Simplify:2sinx cosx (cosx + sinx) / (cosx - sinx).We already know that cosx - sinx = 3√2/5,and we have cosx + sinx = -4√2/5,and 2sinx cosx = 7/25.So, substituting these values in:Numerator: 7/25 * (-4√2/5),Denominator: 3√2/5.So, the expression becomes:(7/25 * -4√2/5) / (3√2/5).Simplify numerator:7/25 * -4√2/5 = (-28√2)/125.Denominator: 3√2/5.So, dividing numerator by denominator:(-28√2)/125 ÷ (3√2)/5 = (-28√2)/125 * 5/(3√2).Simplify:The √2 cancels out,(-28)/125 * 5/3 = (-28*5)/(125*3) = (-140)/375.Simplify the fraction:Divide numerator and denominator by 5: (-28)/75.So, the value is -28/75.Wait, so despite the confusion earlier about cosx being negative, the final answer comes out to -28/75.I think that's correct because even though cosx came out negative, the way the expression is set up, it cancels out appropriately.So, the final answer is -28/75.