find the critical numbers of the function. (enter your answers as a comma - separated list.)\n$h(x)=\\sin…

Explanation:

Step1: Differentiate the function

Using the chain - rule, if (y = \sin^{2}x=( \sin x)^{2}), then (y^\prime = 2\sin x\cos x) (by (u = \sin x), (y = u^{2}), (y^\prime=2u\cdot u^\prime)) and the derivative of (\cos x) is (-\sin x). So (h^\prime(x)=2\sin x\cos x-\sin x=\sin x(2\cos x - 1))

Step2: Set the derivative equal to zero

Set (h^\prime(x)=0), then (\sin x(2\cos x - 1)=0)

• Case 1: (\sin x=0) Since (0\lt x\lt2\pi), then (x = \pi) (because (\sin x = 0) when (x = k\pi,k\in\mathbb{Z}), and in the interval ((0,2\pi)), (k = 1) gives (x=\pi))
• Case 2: (2\cos x-1 = 0) Solve (2\cos x-1 = 0) for (x). First, rewrite it as (\cos x=\frac{1}{2}) Since (\cos x=\frac{1}{2}) and (0\lt x\lt2\pi), then (x=\frac{\pi}{3}) or (x = 2\pi-\frac{\pi}{3}=\frac{5\pi}{3}) (because (\cos x) is positive in the first and fourth quadrants and (\cos\alpha=\cos(2\pi - \alpha)))

Answer:

(\frac{\pi}{3},\pi,\frac{5\pi}{3})