The number of solutions of ${\sin }^{-1}x+{\sin }^{-1}(1-x)={\cos }^{-1}x$ is

THe given inverse trigonometric equation is ${\sin }^{-1}x+{\sin }^{-1}(1-x)={\cos }^{-1}x$

Simplify the above equation as 

$\begin{array}{rcl}{\sin }^{-1}x+{\sin }^{-1}(1-x)& =& {\cos }^{-1}x\\ & =& \frac{\mathrm{\pi }}{2}-{\sin }^{-1}x\\ {\sin }^{-1}\left(1-x\right)& =& \frac{\mathrm{\pi }}{2}-2{\sin }^{-1}x\\ 1-x& =& \cos \left(2{\sin }^{-1}x\right)\\ & =& 1-2x^2\end{array}$

The equation is $2x^2-x=0\Rightarrow x\left(2x-1\right)=0$

Hence, $x=0,\frac{1}{2}$

The number of solutions for the equation ${\sin }^{-1}x+{\sin }^{-1}(1-x)={\cos }^{-1}x$ is $\boxed{\mathbf{2}}$