Problem 9

$\sqrt { 1+\cos { 2x } } =\sqrt { 2 } \sin ^{ -1 }{ \left( \sin { x } \right) }$

$\implies$ $\sqrt{2}|cosx| = \sqrt{2}sin^{-1}(sinx)$

$\implies$ $|cosx| = sin^{-1}(sinx)$ .

When we draw the graph both functions (shown below) we can actually see that they intersect only at two points $\forall~x \in -\pi \le \quad x\le \quad \pi$ .

It was quite amazing that number of solutions are in decimals . :)

$\\ \sqrt{1 + \cos (2x)} = \sqrt{2} \cdot \sin^{- 1} (\sin x) \\\\ \sin^{- 1} (\sin x) = \frac{\sqrt{1 + \cos (2x)}}{\sqrt{2}} \\\\ \sin^{- 1} (\sin x) = \sqrt{\frac{1 + \cos (2x)}{2}} \\\\ \sin^{- 1} (\sin x) = \sqrt{\cos^2 x} \\\\ \sin^{- 1} (\sin x) = \pm \cos x \\\\ \sin (\pm \cos x) = \sin x \Rightarrow \begin{cases} \sin (- cos x) = \sin x \\ \sin (cos x) = \sin x \end{cases}$

Igualando-as,

$\begin{cases} \sin (- cos x) = \sin x \\ \sin (cos x) = \sin x \end{cases} \Rightarrow \begin{cases} - cos x = x \\ cos x = x \end{cases}$

Igualando-as, novamente:

$\\ x = x \\ - \cos x = \cos x \\ 2 \cdot \cos x = 0 \\ \cos x = 0$

Logo, temos que $\boxed{S = \left \{ - \frac{\pi}{2}, \frac{\pi}{2} \right \}}$