Question:

Given: $\cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x + 1)$ Rearranging: $$\cos^{-1} x - \sin^{-1} x - \sin^{-1}(2x + 1) = \pi$$ Since $\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$, we have $\cos^{-1} x - \sin^{-1} x = \pi - 2\sin^{-1} x$ Let $\cos^{-1} x = \alpha$ and $\sin^{-1}(2x + 1) = \beta$ Then: $$2\alpha - \beta = \frac{3\pi}{2}$$ So: $$2\alpha = \frac{3\pi}{2} + \beta$$ Taking cosine of both sides: $$\cos 2\alpha = \cos\left(\frac{3\pi}{2} + \beta\right) = \sin \beta$$ Since $\cos 2\alpha = 2\cos^2 \alpha - 1 = 2x^2 - 1$ and $\sin \beta = 2x + 1$: $$2x^2 - 1 = 2x + 1$$ $$2x^2 - 2x - 2 = 0$$ $$x^2 - x - 1 = 0$$ $$x = \frac{1 \pm \sqrt{5}}{2}$$ Checking domain constraints: only $x = \frac{1 - \sqrt{5}}{2}$ is valid. For $x = \frac{1 - \sqrt{5}}{2}$: $$2x - 1 = 1 - \sqrt{5} - 1 = -\sqrt{5}$$ $$(2x - 1)^2 = 5$$ Therefore: $\sum_{x \in S} (2x - 1)^2 = 5$