Solve $\sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x$ and avoid extra solutions while squaring Solve the equation,

$$
\sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x
$$

My Attempt:
$$
\cos\Big[ \sin^{-1}x+\sin^{-1}(1-x) \Big]=x\\
\cos\big(\sin^{-1}x\big)\cos\big(\sin^{-1}(1-x)\big)-\sin\big(\sin^{-1}x\big)\sin\big(\sin^{-1}(1-x)\big)=x\\
\sqrt{1-x^2}.\sqrt{2x-x^2}-x.(1-x)=x\\
\sqrt{2x-x^2-2x^3+x^4}=2x-x^2\\
\sqrt{x^4-2x^3-x^2+2x}=\sqrt{4x^2-4x^3+x^4}\\
x(2x^2-5x+2)=0\\
\implies x=0\quad or \quad x=2\quad or \quad x=\frac{1}{2}
$$
Actual solutions exclude $x=2$.ie, solutions are $x=0$ or $x=\frac{1}{2}$.
I think additional solutions are added because of the squaring of the term $2x-x^2$ in the steps. 
So, how do you solve it avoiding the extra solutions in similar problems ?
Note: I dont want to substitute the solutions to find the wrong ones.
 A: The domain gives
$$-1\leq x\leq1$$ and $$-1\leq1-x\leq1,$$ which gives $$0\leq x\leq1,$$
which says that the answer is $$\left\{\frac{1}{2},0\right\}.$$
I think it's better after your third step to write
$$\sqrt{2x-x^2}=\sqrt{1-x^2}$$ or $x=0$.
A: Here's a way to avoid the extraneous solution.  Note that $\arcsin u+\arccos u={\pi\over2}$ for all $u\in[-1,1]$. Thus we can rewrite $\arcsin x+\arcsin(1-x)=\arccos x$ as 
$$\arcsin x+{\pi\over2}-\arccos(1-x)={\pi\over2}-\arcsin x$$
which simplifies to
$$2\arcsin x=\arccos(1-x)$$
Applying $\cos$ to each side and using $\cos(2\theta)=1-2\sin^2\theta$, we get $1-2x^2=1-x$, or
$$2x^2-x=0$$
which has $x=0$ and $x={1\over2}$ as its only solutions.
A: Like Barry Cipra,
$$2\arcsin x=\arccos(1-x)$$
Now $0\le\arccos(1-x)\le\pi$ and $-\pi\le2\arcsin x\le\pi$
$\implies\arcsin x\ge0\iff x\ge0$
Now for $\arcsin x\ge0,2\arcsin x=\begin{cases}\arccos(1-2x^2)&\mbox{if }x\ge0\\ 
-\arccos(1-2x^2)& \mbox{if }x<0\end{cases}$
$x\ge0\implies 1-x=1-2x^2$
Can you take it from here?
