Solve the equation: $\arcsin(2x^2 −1) + 2\arcsin x = -\pi/2$ How would I go about solving the equation below:
$\arcsin(2x^2 −1) + 2\arcsin x = -\pi/2$
After appyling sin to both sides I end up with: 
$(2x^2-1)(\sqrt{1-x^2}^2 - x^2) + 2x \sqrt{1 -(2x^2-1)^2}\sqrt{1-x^2}= -1$

Hint:
As $\arcsin(-y)=-\arcsin y$
$$\iff-2\arcsin x=\dfrac\pi2-\arcsin(1-2x^2)=\arccos(1-2x^2)$$
Now $0\le\arccos(1-2x^2)\le\pi\implies0\le-2\arcsin x\le\pi\iff0\ge\arcsin x\ge-\dfrac\pi2$
Now let $\arcsin x=y\implies x=\sin y,0\le-2y\le\pi$
$$\arccos(1-2x^2)=\arccos(\cos2y)=\begin{cases}2y &\mbox{if }0\le2y\le\pi\\
-2y& \mbox{if } 0\le-2y\le\pi \end{cases}$$
Can you take it from here?
