$2 \cos^{-1}x =\sin^{-1}(2x \sqrt {1-x^2})$ is valid for which values of $x $ 
Problem: $2 \cos^{-1}x =\sin^{-1}(2x \sqrt {1-x^2})$ is valid for which values of $x $
Solution:  $2 \cos^{-1}x =\sin^{-1}(2x \sqrt {1-x^2})$
$2 \cos^{-1}x =2 \sin^{-1}x$
$ \cos^{-1}x = \sin^{-1}x$

Am I doing right ?

As the principal value of $\arccos x$ lies in $\in[0,\pi]$
$\implies 2\arccos x$ will lie in $\in[0,2\pi]$
Again as the principal value of $\arcsin x$ $\in[-\frac\pi2,\frac\pi2]$
Observe  that the intersection of region is $[0,\frac\pi2]$
So, $\sin^{-1}(2x \sqrt {1-x^2})=2\arccos x$ 
if $0\le 2\arccos x\le \frac\pi2\iff 0\le \arccos x\le\frac\pi4\iff 1\ge x\ge\cos\frac\pi4=\frac1{\sqrt2} $
