Answer
The theorem applies.
$c=0$
Work Step by Step
The Mean Value Theorem
If $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ ,
then there exists a number $c$ in $(a, b)$ such that
$f^{\prime}(c)=\displaystyle \frac{f(b)-f(a)}{b-a}.$
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f is continuous on the closed interval $[-\pi,\pi]$
$f^{\prime}(x)=1+\sin x$
...and f is differentiable on the open interval $(-\pi,\pi)$
Both premises are satisfied, so the theorem applies.
There is a $ c\in (-\pi,\pi)$ such that
$f^{\prime}(c)=\displaystyle \frac{f(b)-f(a)}{b-a}=\frac{\frac{\pi}{2}-(-\frac{\pi}{2})}{\frac{\pi}{2}-(-\frac{\pi}{2})}=1$
$1+\sin c=1$
$\sin c=0$
$ c=0\in (-\pi,\pi)$
