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

We need to find all solutions of $$\cos^{-1}(2x) - 2\cos^{-1}\!\left(\sqrt{1-x^2}\right) = \pi$$ for $$x \in \left[-\tfrac{1}{2},\,\tfrac{1}{2}\right]$$, and then evaluate $$\displaystyle\sum_{x \in S}\!\left(2\sin^{-1}(x^2) - 1\right)$$.

Let $$A = \cos^{-1}(2x)$$ and $$B = \cos^{-1}\!\left(\sqrt{1-x^2}\right)$$. We require $$A - 2B = \pi$$, i.e., $$A = \pi + 2B$$.

By definition, $$A = \cos^{-1}(2x) \in [0, \pi]$$. Also, for $$x \in [-\tfrac12, \tfrac12]$$ we have $$\sqrt{1-x^2} \in \left[\tfrac{\sqrt{3}}{2},\,1\right]$$, so $$B = \cos^{-1}\!\left(\sqrt{1-x^2}\right) \in \left[0,\,\tfrac{\pi}{6}\right]$$, meaning $$B \geq 0$$.

From $$A = \pi + 2B$$ and $$B \geq 0$$, we get $$A \geq \pi$$. But $$A \leq \pi$$ by the range of $$\cos^{-1}$$. Therefore $$A = \pi$$ and $$B = 0$$ must hold simultaneously.

$$A = \pi$$ requires $$2x = \cos\pi = -1$$, giving $$x = -\tfrac{1}{2}$$.

$$B = 0$$ requires $$\sqrt{1-x^2} = \cos 0 = 1$$, giving $$x = 0$$.

These two conditions are contradictory — no single value of $$x$$ can satisfy both $$x = -\tfrac{1}{2}$$ and $$x = 0$$. Therefore the solution set $$S$$ is empty.

Since $$S = \emptyset$$, the sum over the empty set is $$\displaystyle\sum_{x \in S}\!\left(2\sin^{-1}(x^2) - 1\right) = 0$$.

Hence, the correct answer is Option 1.