Solving $\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$ If we have

$$\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$$

Then, what will be the set of $x$ for which this equation is true?
I tried to solve it by putting $x = \sin a$ or $\cos a$ but got no result.
I am totally stuck on how to do it.

Using  Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
$$
2\arcsin x  =\begin{cases}
\arcsin(2x\sqrt{1-x^2}) \;\;;2x^2\le 1 \\
\pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2 > 1, 0< x\le 1\\
-\pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2> 1, -1< x \le 0
\end{cases}
$$
So the given equation clearly holds true for $2x^2\le1\iff-\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}$
For the other two cases,  we need $$\arcsin(2x\sqrt{1-x^2})=\pm\dfrac\pi2$$
$$\iff2x\sqrt{1-x^2}=\sin\left(\pm\dfrac\pi2\right)=\pm\sin\dfrac\pi2=\pm1$$
$$\iff(2x^2-1)^2=0$$ which has already been covered in the first case
