$f(x) = \sin x - \sin 2x$ in $[0,\pi ]$

We have, $f(x) = \sin x - \sin 2x$ in $[0,\pi ]$

(i) Since, we know that sine functions are continuous functions hence $f(x) = \sin x - \sin 2x$ is a continuous function in $[0,\pi ]$.

(ii) ${f^\prime }(x) = \cos x - \cos 2x \cdot 2 = \cos x - 2\cos 2x,$ which exists in $(0,\pi )$.

So, $f(x)$ is differentiable in $(0,\pi )$. Conditions of mean value theorem are satisfied. Hence, $\exists c \in (0,\pi )$ such that, $f'(c) = \frac{{f(\pi ) - f(0)}}{{\pi - 0}}$

$\Rightarrow$ $\cos c - 2\cos 2c = \frac{{\sin \pi - \sin 2\pi - \sin 0 + \sin 2 \cdot 0}}{{\pi - 0}}$

$\Rightarrow$ $2\cos 2c - \cos c = \frac{0}{\pi }$
$\Rightarrow$ $2 \cdot \left( {2{{\cos }^2}c - 1} \right) - \cos c = 0$

$\Rightarrow$ $4{\cos ^2}c - 2 - \cos c = 0$

$\Rightarrow$ $4{\cos ^2}c - \cos c - 2 = 0$

$\Rightarrow$ $\cos \,c = \frac{{1 \pm \sqrt {1 + 32} }}{8} = \frac{{1 \pm \sqrt {33} }}{8}$

therefore,$c = {\cos ^{ - 1}}\left( {\frac{{1 \pm \sqrt {33} }}{8}} \right)$

Also, ${\cos ^{ - 1}}\left( {\frac{{1 \pm \sqrt {33} }}{8}} \right) \in (0,\pi )$

Hence, mean value theorem has been verified.