<s>[INST] 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: 
$(2x^2-1)(\sqrt{1-x^2}^2 - x^2) + 2x \sqrt{1 -(2x^2-1)^2}\sqrt{1-x^2}= -1$
 [/INST] A method using derivatives: put $f(x) := \arcsin (2x^2-1)+2\arcsin x$. Observe that $f$ is defined on $[-1,1]$, it is continuous and $f(0) = -\pi/2$. Let us compute its derivative:
\begin{align}
f'(x) & = \frac{4x}{\sqrt{1-(2x^2-1)^2}}+\frac{2}{\sqrt{1-x^2}} \\
& = \frac{2x}{\lvert x \rvert \sqrt{1-x^2}}+\frac{2}{\sqrt{1-x^2}}=  
\begin{cases}
0, \text{ if } -1 < x < 0 \\
\frac{4}{\sqrt{1-x^2}} \text{ if } 0 < x < 1.
\end{cases}
\end{align}
So $f$ is constantly $-\pi/2$ on $[-1,0]$. For $x > 0$ its derivative is positive, so $f$ increases strictly and you do not get other solutions to your equation.
A remark on your method: if you decide to apply $\sin$ to both sides be aware that you are actually solving more than one equation, because $-1 = \sin \left(-\pi/2 + k\pi\right)$, with $k \in \mathbb{Z}$. So when you get the final solutions of $\sin(\arcsin(2x^2-1)+2\arcsin x)=-1$ you should check which ones are also solutions of, for example, $\arcsin (2x^2-1)+2\arcsin x = 3\pi/2$. You see, $x=1$ satisfies the latter equation, but not the former. 
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