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

Range of ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$  is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

Range of $2{\cos }^{-1}x$  is $\left[0,2\pi \right]$

If both are equal both should be lies in $\left[0,\frac{\pi }{2}\right]$

$0\le 2{\cos }^{-1}x\le \frac{\pi }{2}\Rightarrow 0\le {\cos }^{-1}x\le \frac{\pi }{4}$

$\frac{1}{\sqrt{2}}\le x\le 1$