Answer
The function has three critical numbers in the specified domain:
{$\frac{\pi}{3}, \pi, \frac{5\pi}{3} $}.
Work Step by Step
$h'(x)=2(\sin{x})(\cos{x})-\sin{x}=\sin{x}(2\cos{x}-1).$
$h'(x)=0\to \sin{x}(2\cos{x}-1)=0\to\sin{x}=0$ or $\cos{x}=\frac{1}{2}.$
$\sin{x}=0\to x=\pi k$ where $k$ is any integer.
$\cos{x}=\frac{1}{2}\to x=\frac{\pi}{3}+2\pi k$ or $x=\frac{5\pi}{3}+2\pi k.$
Plugging in values for $k$ and remembering the limits on x in this problem, we get the following ciritical numbers for $h(x):$
{$\frac{\pi}{3}, \pi, \frac{5\pi}{3} $}
