Question

The number of real solution(s) of the equation $\sqrt{1+\cos }2x=\sqrt{2}{\sin }^{-1}\left(\sin x\right)-\pi \le x\le \pi ,$ is

Answer 1

We have,
$\sqrt{2}|\cos x|=\sqrt{2}{\sin }^{-1}\left(\sin x\right),-\pi \le x\le \pi$
$\Rightarrow |\cos x|=(\pi -x);$
$\frac{\pi }{2}\le x\le \pi \text{and}|\cos x|=x,-\frac{\pi }{2}\le x\le \frac{\pi }{2}$
Clearly $y=|\cos x|\text{and}y=(\pi -x)y=x$ intersect at two distinct points.