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?
 A: In your first step you added an extra solution. 
Since $\arcsin x$ must be smaller than $\pi/2$, the first line reads:
$$\arcsin(1-x)= \frac{\pi}{2}+2\arcsin(x) \le \frac{\pi}{2}$$
Thus, $x\le 0$ as well.
Now, by taking the $\sin$ of both sides, you took a function that was only defined up to $x=1$ (e.g. $\arcsin(x-1)$ ) and extended it to all reals (e.g $x-1$). Here is where you added the extra solution.
A: 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 [$0<x<1$]).
A: As  $\displaystyle \sin\left(\frac\pi2\pm A\right)=\cos A,$
$\displaystyle\sin\left(\frac\pi2\pm2\arcsin x\right)=\cos(2\arcsin x)=1-2\left[\sin(\arcsin x)\right]^2=1-2x^2$
So, $\displaystyle x=\frac12$ corresponds to $\displaystyle\arcsin(1-x)+2\arcsin x=\frac\pi2$ as $\displaystyle\arcsin\frac12=\frac\pi6$
