<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] Consider the function
$$
f(x)=\arcsin(2x\sqrt{1-x^2})-2\arcsin x
$$
which is defined in the domain where
$$
\begin{cases}
|2x\sqrt{1-x^2}|\le 1\\[4px]
|x|\le 1
\end{cases}
$$
The first inequality becomes, writing $t=|x^2|$,
$$
4t-4t^2\le 1
$$
that's satisfied for every $t$. So our function $f$ is defined over $[-1,1]$.
The derivative of $f$ is
$$
f'(x)=\frac{1}{\sqrt{1-4x^2(1-x^2)}}\left(2\sqrt{1-x^2}-\frac{2x^2}{\sqrt{1-x^2}}\right)-\frac{2}{\sqrt{1-x^2}}
$$
that can be simplified to
$$
f'(x)=\frac{2}{\sqrt{1-x^2}}
\left(\frac{1-2x^2}{|1-2x^2|}-1\right)
$$
The derivative exists in the set
$$
(-1,-1/\sqrt{2})\cup(-1/\sqrt{2},1/\sqrt{2})\cup(1/\sqrt{2},1)
$$
and it can be simplified further as
$$
f'(x)=\begin{cases}
-\dfrac{4}{\sqrt{1-x^2}} & \text{for $1/\sqrt{2}<|x|<1$}\\[8px]
0 & \text{for $|x|<1\sqrt{2}$}
\end{cases}
$$
Thus the function is constant over $[-1/\sqrt{2},1/\sqrt{2}]$ (also at the extremes, by continuity). Since $f(0)=0$, we have that the given identity is valid on this interval and nowhere else, because over $[-1,-1/\sqrt{2}]$ and $[1/\sqrt{2},1]$ the function is strictly decreasing.

</s>