Value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$ How can we find the value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$? 
Note that $\sin^{-1}$ is the inverse sine function.
I'm asking for the solution $x$ for this equation.
Please work out the solution.

Let $\sin^{-1}x=y\implies-\dfrac\pi2\le y\le\dfrac\pi2$
$\implies y+\sin^{-1}(1-\sin y)=\dfrac\pi2-y$
$\implies\sin^{-1}(1-\sin y)=\dfrac\pi2-2y$
$\implies1-\sin y=\sin\left(\dfrac\pi2-2y\right)=\cos2y=1-2\sin^2y$
$\implies2\sin^2y-\sin y=0$
Observation :
For real $\sin^{-1}x, -1\le x\le1$
and for real $\sin^{-1}(1-x),-1\le1-x\le1\implies2\ge x\ge0$
$\implies0\le x\le1\implies0\le y\le\dfrac\pi2$
