\(2 \cos ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)\) is valid for all values of \(x\) satisfying

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
\[
\begin{array}{r}
0 \leq \mathrm{y} \leq \frac{\pi}{4} \Rightarrow 1 \geq \cos \mathrm{y} \geq \frac{1}{\sqrt{2}} \\
\Rightarrow \quad \frac{1}{\sqrt{2}} \leq \mathrm{x} \leq 1 \Rightarrow x \in\left[\frac{1}{\sqrt{2}}, 1\right]
\end{array}
\]