Answer
Rolle's theorem can be applied; $c=\dfrac{\pi}{2}$ or $c=\dfrac{3\pi}{2}.$
Work Step by Step
$f(x)$ is defined for all values of $x$ and is differentiable at every value of $x.$
$f(0)=f(2\pi)=0.$
Since $f(x)$ is continuous over $[0 , 2\pi]$ and differentiable over $(0, 2\pi)$, applying Rolle's Theorem over the interval $[0, 2\pi]$ guarantees the existence of at least one value $c$ such that $0\lt c\lt 2\pi$ and $f'(c)=0.$
$f'(x)=\cos{x}.$
$f'(x)=0\to \cos{x}=0\to c=\dfrac{\pi}{2}$ or $c=\dfrac{3\pi}{2}.$
