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?

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.
