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Find all critical points of the function \newlineg(θ)=sin⁡2(6θ)g(\theta)=\sin^{2}(6\theta) Express your answer in terms of π\pi. \newline(Use symbolic notation and fractions where needed. Use nn for all integer values.)\newlineθ=\theta=

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Q. Find all critical points of the function \newlineg(θ)=sin⁡2(6θ)g(\theta)=\sin^{2}(6\theta) Express your answer in terms of π\pi. \newline(Use symbolic notation and fractions where needed. Use nn for all integer values.)\newlineθ=\theta=
  1. Find Derivative of g(θ)g(\theta): To find the critical points of the function g(θ)=sin⁡2(6θ)g(\theta) = \sin^2(6\theta), we need to find the values of θ\theta where the derivative of gg with respect to θ\theta is zero or undefined.\newlineFirst, let's find the derivative of g(θ)g(\theta).\newlineUsing the chain rule, the derivative of sin⁡2(u)\sin^2(u) with respect to uu is 2sin⁡(u)cos⁡(u)2\sin(u)\cos(u), where u=6θu = 6\theta in our case.\newlineSo, the derivative of g(θ)g(\theta) with respect to θ\theta is g(θ)=sin⁡2(6θ)g(\theta) = \sin^2(6\theta)22.\newlineSince g(θ)=sin⁡2(6θ)g(\theta) = \sin^2(6\theta)33, we have g(θ)=sin⁡2(6θ)g(\theta) = \sin^2(6\theta)44.\newlineSimplify this to get g(θ)=sin⁡2(6θ)g(\theta) = \sin^2(6\theta)55.
  2. Find g′(θ)=0g'(\theta) = 0: Now, we need to find the values of θ\theta where g′(θ)=0g'(\theta) = 0. The derivative g′(θ)=12sin⁡(6θ)cos⁡(6θ)g'(\theta) = 12\sin(6\theta)\cos(6\theta) is zero when either sin⁡(6θ)=0\sin(6\theta) = 0 or cos⁡(6θ)=0\cos(6\theta) = 0. Let's solve for θ\theta when sin⁡(6θ)=0\sin(6\theta) = 0 first. sin⁡(6θ)=0\sin(6\theta) = 0 when 6θ=nπ6\theta = n\pi, where θ\theta00 is an integer. Divide both sides by θ\theta11 to get θ\theta22.
  3. Solve for sin⁡(6θ)=0\sin(6\theta) = 0: Next, let's solve for θ\theta when cos⁡(6θ)=0\cos(6\theta) = 0. cos⁡(6θ)=0\cos(6\theta) = 0 when 6θ=(2n+1)π26\theta = (2n+1)\frac{\pi}{2}, where nn is an integer. Divide both sides by 66 to get θ=(2n+1)π12\theta = \frac{(2n+1)\pi}{12}.
  4. Solve for cos⁡(6θ)=0\cos(6\theta) = 0: We have two sets of solutions for θ\theta: θ=nπ/6\theta = n\pi/6 and θ=(2n+1)π/12\theta = (2n+1)\pi/12. However, we notice that the solutions where θ=(2n+1)π/12\theta = (2n+1)\pi/12 are already included in the solutions where θ=nπ/6\theta = n\pi/6, because when nn is even, (2n+1)π/12(2n+1)\pi/12 simplifies to (mπ)/6(m\pi)/6 where mm is an integer. Therefore, we only need to consider the solutions θ=nπ/6\theta = n\pi/6.
  5. Combine Solutions: The critical points of the function g(θ)g(\theta) are the values of θ\theta where the derivative is zero, which we found to be θ=nπ/6\theta = n\pi/6. These are all the critical points of the function g(θ)=sin⁡2(6θ)g(\theta) = \sin^2(6\theta).

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