Answer
Rolle's Theorem can be applied; $c$ belongs to $\{-\frac{\pi}{2}, 0, \frac{\pi}{2}\}.$
Work Step by Step
$f(x)$ is continuous for all values of $x$ and is differentiable at every value of $x.$
$f(-\pi)=f(\pi)=1.$
Since $f(x)$ is continuous over $[-\pi , \pi]$ and differentiable over $(-\pi, \pi)$, applying Rolle's Theorem over the interval $[-\pi, \pi]$ guarantees the existence of at least one value $c$ such that $-\pi\lt c\lt \pi$ and $f'(c)=0.$
$f'(x)=-2\sin{2x}.$
$f'(x)=0\to -2\sin{2x}=0\to 2x=\pi k\to x=\frac{k}{2}\pi$ where k is any integer.
Substituting values for $k$ and checking the specified interval gives us the following solution set: $\{-\frac{\pi}{2}, 0, \frac{\pi}{2}\}.$
