Solve $\arcsin{(2x^2-1)}+2\arcsin{x} = -\frac{\pi}{2}$ First of, we have to restrict the domain of the equation by looking at the argument of the first term. The domain for $\arcsin$ is $[-1,1]$, so the inequality $ -1\leq 2x^2-1 \leq1$ has to hold. So $x\geq0$ and $-1\leq x\leq 1$. It follow that the domain is $[-1,1]\cap[0,\infty)=[0,1].$ So the real domain of the equation remains $[0,1]$ and if there exists $x\in\mathbb{R}$ that solves the equation it should belong to that domain.
Lets solve it by rewriting $$\arcsin{(2x^2-1)}=-\frac{\pi}{2}-2\arcsin{x}=-\left(\frac{\pi}{2}+2\arcsin{x}\right).$$
Sine of both sides gives $$2x^2-1=\sin\left(\frac{\pi}{2}+2\arcsin{x}\right)=\cos{(2\arcsin{x})}=\cos^2{(\arcsin{x})}-\sin^2{(\arcsin{x})}.$$
Using $\cos^2{x}=1-\sin^2{x}$ we get
$2x^2-1=1-2\sin^2(\arcsin{x})=1-2x^2 \Longleftrightarrow2x^2-1\Longleftrightarrow x_{1,2}=\pm\frac{1}{\sqrt{2}}.$
Only $x=\frac{1}{\sqrt{2}}\in{[0,1]},$ which is the only root.
According to wolfram alpha, the only real solutions are $x=0$ and $x=-1.$ Where did I go wrong?

Hint :
Let $\arcsin x=u\implies-\dfrac\pi2\le u\le\dfrac\pi2\iff-\pi\le2u\le\pi$ and $x=\sin u$
$\arcsin(2x^2-1)=\arcsin(-\cos2u)=-\arcsin(\cos2u)=\arccos(\cos2u)-\dfrac\pi2$
Now $\arccos(\cos2u)=\begin{cases}2u &\mbox{if } 0\le2u\le\pi \\-2u& \mbox{if } 0\le-2u\le\pi\iff-\pi\le 2u\le0&\\2\pi-2u& \mbox{if } 0\le2\pi-2u\le\pi\iff\pi\le 2u\le2\pi\\2\pi+2u& \mbox{if } 0\le2\pi+2u\le\pi\iff-2\pi\le2u\le-\pi\end{cases}$
