Let $\text{S}$  be the set of all solutions of the equation ${\cos }^{-1}\left(2x\right)-2{\cos }^{-1}\left(\sqrt{1-x^2}\right)=\pi ,x\in [-\frac{1}{2},\frac{1}{2}]$ Then  $\sum _{\mathbf{x}\in \mathbf{S}}2{\sin }^{-1}(x^2-1)$ is equal to

 ${\cos }^{-1}\left(2x\right)-2{\cos }^{-1}\sqrt{1-x^2}=\pi$
${\cos }^{-1}\left(2x\right)-{\cos }^{-1}(2(1-x^2)-1)=\pi$
${\cos }^{-1}\left(2x\right)-{\cos }^{-1}(1-2x^2)=\pi$
$-{\cos }^{-1}(1-2x^2)=\pi -{\cos }^{-1}\left(2x\right)$
Taking  $\cos$ both sides we get
$\mathrm{Cos}(-{\cos }^{-1}(1-2x^2))=\cos \left(\pi -{\cos }^{-1}\left(2x\right)\right)$
$1-2x^2=-2x$
$2x^2-2x-1=0$
On solving,  $x=\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2}$
As $x=[-1/2,1/2],x=\frac{1+\sqrt{3}}{2}=$  rejected
So  $x=\frac{1-\sqrt{3}}{2}\Rightarrow x^2-1=-\sqrt{3}/2$
$=2{\sin }^{-1}(x^2-1)=2{\sin }^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\frac{-2\pi }{3}$