Solve the equation $2\arcsin x=\arcsin(\frac{3}{4}x)$ $$2\arcsin x=\arcsin(\frac{3}{4}x)$$ so $x\in[-1,1]$
so we have:
$2\arcsin x=y\Rightarrow\sin\frac{y}{2}=x$ and $\arcsin x=y \Rightarrow \sin y=\frac{3}{4}x\Rightarrow\frac{4}{3}\sin y=x$ ,
$y\in[-\frac{\pi}{2},\frac{\pi}{2}]$
$$\sin\frac{y}{2}-\frac{4}{3}\sin y=0$$$$\sin\frac{y}{2}-\frac{4}{3}\cdot 2\sin\frac{y}{2}\cos\frac{y}{2}=0$$ $$\sin\frac{y}{2}\cdot (1-\frac{8}{3}\cos\frac{y}{2})=0$$
Is it done properly at this point?
1.$\sin \frac{y}{2}=0\Rightarrow x=0$
And what else?
 A: $\sin\frac{y}{2}\cdot (1-\frac{8}{3}\cos\frac{y}{2})=0$
$\sin\frac{y}{2} = 0$ 
So far so good 
contining... suppose $\sin\frac{y}{2} \ne 0$
$(1-\frac{8}{3}\cos\frac{y}{2})=0\\
\frac{8}{3}\cos\frac{y}{2})=1\\
\cos\frac{y}{2}=\frac 38\\
\frac y2=\arccos\frac 38\\
x = \sin(\arccos\frac 38)\\
x = \sqrt {1 - \frac {9}{64}}\\
x = \frac {\sqrt{55}}{8}$
However, if this were the case.
since $\frac {\sqrt 2}{2}< \frac {\sqrt {55}}{8}< 1 \implies \frac {\pi}{4} < \arcsin \frac {\sqrt {55}}{8}< \frac {\pi}{2}\\2\arcsin \frac {\sqrt {55}}{8} > \frac {\pi}{2}$ 
while 
$\arcsin \frac 34 \frac {\sqrt {55}}{8} < \frac {\pi}{2}$
$x = 0$ is the only solution.
One more consideration.  Had we found a non-zero solution, then $-x$ would also be a solution, and the technique here managed to obliterate that possibility.
Cutting out all of the algebra:
$\forall x\ne 0, |2\arcsin x| > |\arcsin(\frac 34 x)|$
Which means that $2\arcsin x \ne \arcsin(\frac 34 x)$
unless $x = 0$
A: If $2 \arcsin(x) = \arcsin(3x/4)$, we apply $\sin$ to both sides, and use the double-angle formula and the fact that $\cos(\arcsin(x)) = \sqrt{1-x^2}$ to get
$$ 2 x \sqrt{1-x^2} = 3x/4 $$
thus either $x=0$ or $\sqrt{1-x^2} = 3/8$.  Squaring both sides of the latter, 
$ 1 - x^2 = 9/64$, so $x = \pm \sqrt{55}/8$.  Note that each of these is in the correct interval for $\arcsin(x)$ and $\arcsin(3x/4)$ to be defined.  But they are not solutions, because $2 \arcsin(\pm \sqrt{55}/8) \approx 2.372799104$ is outside of the interval $[-\pi/2, \pi/2]$ of allowed values for arcsin.
So the only correct solution is $x=0$.
A: HINT:
Take the sine of both sides, apply the double angle formula for the sine function, and use $\cos(\arcsin(x))=\sqrt{1-x^2}$.
