The given function $f(x)=\cos x-\sin 2 x$, is a
continuous function in interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
and differentiable in interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Now, according to Lagrange's mean value theorem, there exist
$c \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that
$f'(c)=\frac{f\left(\frac{\pi}{2}\right)-f\left(-\frac{\pi}{2}\right)}{\frac{\pi}{2}- \left(-\frac{\pi}{2}\right)}$
$\Rightarrow -\sin c-2 \cos (2 c)=0$
$\Rightarrow \sin c+2-4 \sin ^{2} c=0$
$\Rightarrow 4 \sin ^{2} c-\sin c-2=0$
So, $\sin c=\frac{1 \pm \sqrt{1+32}}{8}$
$\Rightarrow \sin c=\frac{1 \pm \sqrt{33}}{8}$
$\Rightarrow c=\sin ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right)$