Put $cos^{-1}\, x = y, so that $x = cos\,y$
Then , $0 \,\le$ y \,\le\,\pi$ and $\left|x\right| \le \,1 \,\,\,\,\,\dots(i)$
and the RHS of given equation becomes
$sin^{-1} \left(2\,cos \,y \,sin \,y\right)=sin^{-1} \left(sin\, 2y\right)=2y$
Since, $sin^{-1} \left(2x \sqrt{1-x^{2}}\right)$ lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}.$
$\therefore $ 2y lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.
i.e., y lies between - $\frac{\pi}{4}$ and $\frac{\pi}{4}$.
$\therefore -\frac{\pi}{4}\le y\le\frac{\pi}{4}.\,\,\,\,\,\dots(ii)$
On combining Eqs. (i) and (ii), we get
$0\le\, y\le\,\frac{\pi}{4} $
$\Rightarrow 1\ge cos \,y \ge\frac{1}{\sqrt{2}}$
$\Rightarrow \frac{1}{\sqrt{2}} \le x \le 1 $
$\Rightarrow x \in\left[\frac{1}{\sqrt{2}}, 1\right]$