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(B) Given equation: $\cos ^{-1}(2 x)-2 \cos ^{-1} \sqrt{1-x^2}=\pi$.Since $x \in [-\frac{1}{2}, \frac{1}{2}]$,let $x = \sin 	heta$,where $	heta \in

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$-\frac{2 \pi}{3}$

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(B) Given equation: $\cos ^{-1}(2 x)-2 \cos ^{-1} \sqrt{1-x^2}=\pi$.Since $x \in [-\frac{1}{2}, \frac{1}{2}]$,let $x = \sin 	heta$,where $	heta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$.Then $\sqrt{1-x^2} = \cos 	heta$. Since $	heta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$,$\cos 	heta \geq 0$,so $\cos^{-1}(\sqrt{1-x^2}) = \cos^{-1}(\cos 	heta) = |	heta|$.Since $	heta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$,$|	heta| = 	heta$ if $	heta \geq 0$ and $-	heta$ if $	heta Case $1$: $x \geq 0 \implies 	heta \in [0, \frac{\pi}{6}]$. The equation becomes $\cos^{-1}(2x) - 2	heta = \pi \implies \cos^{-1}(2x) = \pi + 2	heta$. This is impossible as the range of $\cos^{-1}$ is $[0, \pi]$.Case $2$: $x The equation becomes $\cos^{-1}(2x) - 2\cos^{-1}(\sqrt{1-x^2}) = \pi$.Using $\cos^{-1}(\sqrt{1-x^2}) = \sin^{-1}(|x|) = \sin^{-1}(-x)$,we have $\cos^{-1}(2x) - 2\sin^{-1}(-x) = \pi$.Since $\cos^{-1}(2x) = \frac{\pi}{2} - \sin^{-1}(2x)$,we get $\frac{\pi}{2} - \sin^{-1}(2x) + 2\sin^{-1}(x) = \pi \implies 2\sin^{-1}(x) - \sin^{-1}(2x) = \frac{\pi}{2}$.Testing $x = -\frac{1}{2}$: $\cos^{-1}(-1) - 2\cos^{-1}(\sqrt{3}/2) = \pi - 2(\pi/6) = \pi - \pi/3 = 2\pi/3 
eq \pi$.Solving $2x^2 - 2x - 1 = 0$ gives $x = \frac{1 \pm \sqrt{3}}{2}$. Only $x = \frac{1-\sqrt{3}}{2}$ is in $[-\frac{1}{2}, \frac{1}{2}]$.For $x = \frac{1-\sqrt{3}}{2}$,$x^2 - 1 = (\frac{1-\sqrt{3}}{2})^2 - 1 = \frac{1+3-2\sqrt{3}}{4} - 1 = \frac{4-2\sqrt{3}}{4} - 1 = 1 - \frac{\sqrt{3}}{2} - 1 = -\frac{\sqrt{3}}{2}$.Thus,$2\sin^{-1}(x^2-1) = 2\sin^{-1}(-\frac{\sqrt{3}}{2}) = 2(-\frac{\pi}{3}) = -\frac{2\pi}{3}$.

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

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