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?

Let $f(x)=\arcsin{(2x^2-1)}+2\arcsin{x}+\frac{\pi}{2}$.
Thus, $$f'(x)=\frac{2x}{|x|\sqrt{1-x^2}}+\frac{2}{\sqrt{1-x^2}}=0$$
for all $-1<x<0$.
Thus, $f(x)=conctant$ for $-1\leq x\leq0$ and since $f(0)=0$ we obtain $[-1,0]$.
In another hand $f'(x)>0$ for all $0<x<1$.
Thus, $f(x)>f(0)=0$ and on $[-1,0]$ we got all solutions.
