Let $S=\{x:\cos^{-1}x=\pi+\sin^{-1}x+\sin^{-1}(2x+1)\}$. Then $\displaystyle\sum_{x\in S}(2x-1)^2$ is equal to ________.
Step-by-Step Solution
Key Concept: Rearrange to $2\cos^{-1}x-\sin^{-1}(2x+1)=\tfrac{3\pi}{2}$, then take cosine of both sides using $\cos(2\alpha)=\sin(\beta)$ to get an algebraic quadratic, and discard roots outside the domain.
Rearranging: $\cos^{-1}x-\sin^{-1}x-\sin^{-1}(2x+1)=\pi \Rightarrow 2\cos^{-1}x-\sin^{-1}(2x+1)=\dfrac{3\pi}{2}$.
Let $\alpha=\cos^{-1}x$, $\beta=\sin^{-1}(2x+1)$. Then $2\alpha=\dfrac{3\pi}{2}+\beta$, so $\cos 2\alpha=\sin\beta$.
$2x^2-1=2x+1 \Rightarrow x^2-x-1=0 \Rightarrow x=\dfrac{1\pm\sqrt{5}}{2}$.
Only $x=\dfrac{1-\sqrt{5}}{2}$ is valid (domain constraint).
$(2x-1)^2$: from $x^2-x-1=0 \Rightarrow 4x^2-4x=4 \Rightarrow (2x-1)^2=5$.
$\displaystyle\sum_{x\in S}(2x-1)^2=5$.
Correct Answer: 5