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.
 A: 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 $|x|\le 1/\sqrt 2$.
A: This is essentially the same approach as in the answers of Doug M and Dr. MV, just written to separate the "if" from the "only if" in showing that $\arcsin(2x\sqrt{1-x^2})=2\arcsin x$ if and only if $-1/\sqrt2\le x\le1/\sqrt2$.
If $\arcsin(2x\sqrt{1-x^2})=2\arcsin x$, then the fact that range of the arcsine function is $[-\pi/2,\pi/2]$ tells us 
$$-{\pi\over2}\le\arcsin(2x\sqrt{1-x^2})\le{\pi\over2}\implies-{\pi\over4}\le\arcsin x\le{\pi\over4}\implies-{1\over\sqrt2}\le x\le{1\over\sqrt2}$$
On the other hand, if $-1/\sqrt2\le x\le1/\sqrt2$, then $x$ can be written as $\sin\theta$ with $-\pi/4\le\theta\le\pi/4$.  For such $\theta$'s, we have $\cos\theta=\sqrt{1-\sin^2\theta}$, $\arcsin(\sin\theta)=\theta$, and $\arcsin(\sin2\theta)=2\theta$, and it follows that
$$\arcsin(2x\sqrt{1-x^2})=\arcsin(2\sin\theta\cos\theta)=\arcsin(\sin2\theta)=2\theta=2\arcsin(\sin\theta)=2\arcsin x$$
Thus the solution set is precisely the interval $-1/\sqrt2\le x\le1/\sqrt2$.
A: suppose $x = \sin a$:
$\arcsin(2\sin a \cos a) = 2 \sin a$
$\arcsin(\sin 2a) = 2a$
The range of $\arcsin$ is $[-\pi/2, \pi/2]$
$\arcsin(\sin 2a) = 2a$ is true when $2a$ is within that range.
$a$ in $[-\pi/4,\pi/4]$
$x$ in $[\sin (-\pi/4), \sin(\pi/4)]$
A: 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.

A: 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
