How to solve :$\operatorname{arcsin}x-\operatorname{arcsin}\left(x/2\right)=\operatorname{arcsin}\left(\frac{x\sqrt3}2\right)$ $$\operatorname{arcsin}x-\operatorname{arcsin}\left(x/2\right)=\operatorname{arcsin}\left(\frac{x\sqrt3}2\right)$$
Why can't I figure this one out??
Is it possible to cancel out the arcsins?
I know from graphing on my calculator that the answers are $-1, 0$, and $1$, but want help getting there by hand.
Thank you!
 A: this makes sense ony if $|x|<1$
\begin{split}
&&\arcsin x - \arcsin \dfrac x 2 = \arcsin \dfrac{x\sqrt3}2 \\ 
&&\sin\left(\arcsin x - \arcsin \dfrac x 2\right) =\sin\arcsin \dfrac{x\sqrt3}2 \\ 
&&\sin(\arcsin x )\cos(\arcsin \frac x 2)- \cos(\arcsin x )\sin (\arcsin \dfrac x 2) = \frac{x\sqrt3}{2} \\ 
&&x\sqrt{1-\left(\dfrac x2\right)^2} - \dfrac x2 \sqrt{1-x^2} =\dfrac{x\sqrt3}2 \\ 
&&x\sqrt{4-x^2} - x \sqrt{1-x^2} = x\sqrt3 \\ 
&& x^2\left(5-2x^2 - 2 \sqrt{(4-x^2)(1-x^2)}\right)= 3x^2 \\ 
\end{split}
Hence $$x= 0 \qquad or \qquad 5-2x^2 - 2 \sqrt{(4-x^2)(1-x^2) } = 3$$
However, 
\begin{split}
&&5-2x^2 - 2 \sqrt{(4-x^2)(1-x^2) } = 3\\
&\implies &  \sqrt{(4-x^2)(1-x^2) } = 1-x^2\\
&\implies & \sqrt{(4-x^2) } =\sqrt{(1-x^2) } \qquad or \qquad 1-x^2 = 0
\end{split}
But $ \sqrt{(4-x^2) } =\sqrt{(1-x^2) }$ is impossible

Conclusion $x\in \{-1,0,1\}$ is the set of solutions

A: Use $\sin(A-B)$ identity and the fact that $f(f^{-1}(x)) = x$:
$$x\sqrt{1-\frac{x^2}{4}}-\frac{x}{2}\sqrt{1-x^2} = \frac{x\sqrt 3}{2}$$
So $x=0$ is one solution. Now 
$$\sqrt{4-x^2}-\sqrt{1-x^2} = \sqrt{3}$$
Now let $1-x^2 = t$, so that our equation becomes:
$$\sqrt{3+t}-\sqrt{t} = \sqrt{3}$$
Upon squaring, we get:
$$3+t+t - 2\sqrt{(3+t)t} = 3 \\
t^2 = (3+t)(t)$$
From here, we get $t = 0$. Thus $1-x^2 = 0$, so that $x = \pm 1$
So we have three solutions, $x = -1, 0, 1$, which satisfy the original equation.
