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

$$\begin{gathered} \begin{array}{l}{\cos }^{-1}\left(2\mathrm{x}\right)-2{\cos }^{-1}\left(\sqrt{1-{\mathrm{x}}^2}\right)=\mathrm{\pi }\\ \Rightarrow 2\mathrm{x}=-\cos \left(2{\cos }^{-1}\sqrt{1-{\mathrm{x}}^2}\right) \\ x=\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2} \\ \text{ As }x=[-1/2,1/2],x=\frac{1+\sqrt{3}}{2}=\text{ rejected }\\ \Rightarrow \mathrm{x}=\frac{1-\sqrt{3}}{2} \\ \Rightarrow x^2-1=-\sqrt{3}/2\\ \sum _{\mathrm{x}\in \mathrm{S}} 2{\sin }^{-1}\left({\mathrm{x}}^2-1\right)=\frac{-2\mathrm{\pi }}{3}\end{array} \end{gathered}$$