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.

My answer is similar to @juantheron.
First we know that the equation is valid for $0\le\theta\le\dfrac{\pi}2,$ for some $\theta$ in a right triangle.  Now taking the cosine of both sides we have:
$$x=\cos(\sin^{-1}(x)+\sin^{-1}(1-x).$$
Using the angle sum formula for cosine and the Pythagorean Theorem, we get
\begin{align}
x&=\cos(\sin^{-1}(x))\cos(\sin^{-1}(1-x))-\sin(\sin^{-1}(x))\sin(\sin^{-1}(1-x))\\
&=\sqrt{1-x^2}\sqrt{1-(1-x)^2}-x(1-x)\\
&=\sqrt{(1-x^2)(2x-x^2)}-x(1-x)\\
2x-x^2&=\sqrt{(1-x^2)(2x-x^2)}\\
\end{align}
Now, noting that $x=0$ is a solution, the remaining solutions come from
\begin{align}
\sqrt{2x-x^2}&=\sqrt{1-x^2}\\
2x&=1\\
x&=\dfrac12.
\end{align}
