$2 \cos^{-1}x =\sin^{-1}(2x \sqrt {1-x^2})$ 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 ?
 A: Essentially, we are looking for the solutions to:
$$2\cos^{-1}(x)=\sin^{-1}\Big(2x\sqrt{1 - x^2}\Big)$$
This is the sort of problem where a graph can help you gain intuition with respect to the interval over which the equation is valid:
See, e.g., WolframAlpha:

As you can see from the graphs of $\color{blue}{\bf 2\cos^{-1}(x)}\,$ and $\,\color{purple}{\bf \sin^{-1}(2x\sqrt{1 - x^2})}\,$ above, the two curves coincide on the interval $\left[\dfrac{\sqrt 2}2, 1\right]$.
You can confirm, algebraically, that the solutions to the given equation consists of those values of x at which the two graphs intersect, so that the interval over which equality holds is $$\dfrac {\sqrt 2}2 \leq x \leq 1.$$
A: 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} $
