<s>[INST] 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.
 [/INST] Note that if 
$$\arcsin(2x\sqrt{1-x^2})=2\arcsin(x) \tag 1$$
then taking the sine of both sides of $(1)$ yields
$$\begin{align}
2x\sqrt{1-x^2}&=\sin(2\arcsin(x))\\\\
&=2\sin(\arcsin(x))\cos(\arcsin(x))\\\\
&=2x\sqrt{1-x^2}
\end{align}$$
However, taking the cosine of the left-hand side of $(1)$ yields
$$\cos(\arcsin(2x\sqrt{1-x^2}))=|1-2x^2|$$
while taking the cosine of the right-hand side of $(1)$ yields
$$\cos(2\arcsin(x))=1-2x^2$$
Therefore, the equality 
$$\arcsin(2x\sqrt{1-x^2})=2\arcsin(x)$$
is valid for $|x|\le 1/\sqrt 2$.
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