Let \( S \) be the set of all solutions of the equation \( \cos^{-1}(2x) - 2\cos^{-1}\!\left(\sqrt{1-x^2}\right) = \pi , \; x \in \left[ -\frac{1}{2} , \frac{1}{2} \right] .\) Then \( \displaystyle \sum_{x \in S} 2\sin^{-1}(x^2-1) \) is equal to:
\( 0 \)
\( \pi - 2\sin^{-1}\!\left(\frac{\sqrt{3}}{4}\right) \)
\( -\frac{2\pi}{3} \)
\( \pi - \sin^{-1}\!\left(\frac{\sqrt{3}}{4}\right) \)
