Answer
Rolle's Theorem can be applied; $c=\pi$.
Work Step by Step
$f(x)$ is continuous for all values of $x$ and is differentiable at every value of $x.$
$f(0)=f(2\pi)=1.$
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 $f'(c)=0.$
$f'(x)=-\sin{x}.$
$f'(x)=0\to -\sin{x}=0\to c=\pi k$ where $k$ is any integer.
Substituting values for $k$ gives us that $c=\pi$.
