<s>[INST] 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!
 [/INST] 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

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