Answer
The Mean Value Theorem can be applied; $c=\dfrac{\pi}{2}.$
Work Step by Step
$f(x)$ is continuous for all values of $x$ and differentiable at every value of $x.$
$a=0; b=\pi\to f(a)=0$ and $f(b)=0.$
Since $f(x)$ is continuous over $[a, b]$ and differentiable over $(a, b)$, applying Mean Value Theorem over the interval $[a , b]$ guarantees the existence of at least one value c such that $a\lt c\lt b$ and $fâ˛(c)=\dfrac{f(b)-f(a)}{b-a}.$
$f'(c)=\dfrac{0-0}{\pi-0}=0.$
$f'(x)=\cos{x}\to f'(c)=\cos{c}=0\to c=\dfrac{\pi}{2}.$
