Solving $\arcsin(1-x)-2\arcsin(x)=\pi/2$ \begin{eqnarray*}
\arcsin(1-x)-2\arcsin(x) & = & \frac{\pi}{2}\\
1-x & = & \sin\left(\frac{\pi}{2}+2\arcsin(x)\right)\\
 & = & \cos\left(2\arcsin(x)\right)\\
 & = & 1-2\left(\sin\left(\arcsin(x)\right)\right)^{2}\\
 & = & 1-2x^{2}\\
x & = & 2x^{2}\\
x\left(x-\frac{1}{2}\right) & = & 0
\end{eqnarray*}
So $x=0$ or $x=\frac{1}{2}$
But puttig $x=\frac{1}{2}$ in the original expression gives $-\frac {\pi} 4 \ne \frac \pi 2$
So, why do we get $x=-1/2$ as an answer?

Beside the good answers you already received, you can also consider the problem from an algebraic point of view. Let $$f(x)=\arcsin(1-x)-2\arcsin(x)-\frac{\pi }{2}$$ Its derivative $$f'(x)=-\frac{2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{(2-x) x}}$$ is always negative (with infinite branches at $x=0$ and $x=1$). For $x=0$, $f(0)=0$ and since the function decreases, you cannot have any root beside $x=0$ (remember that $f(x)$ is only defined for [$0<x<1$]).
