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