Let $\mathrm{S}=\left\{x: \cos ^{-1} x=\pi+\sin ^{-1} x+\sin ^{-1}(2 x+1)\right\}$. Then $\sum_{x \in \mathrm{~S}}(2 x-1)^2$ is equal to ______.

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Aicharya · Accepted Answer

$\begin{aligned} & \cos ^{-1} x=\pi+\sin ^{-1} x+\sin ^{-1}(2 x+1) \\ & 2 \cos ^{-1} x-\sin ^{-1}(2 x+1)=\frac{3 \pi}{2} \\ & 2 \alpha-\beta=\frac{3 \pi}{2} \text { where } \cos ^{-1} x=\alpha, \sin ^{-1}(2 x+1)=\beta \\ & 2 \alpha=\frac{3 \pi}{2}+\beta \\ & \cos 2 \alpha=\sin \beta \\ & 2 \cos ^2 \alpha-1=\sin \beta \\ & 2 x^2-1=2 x+1 \\ & x^2-x-1=0\end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{n}=\frac{1 \pm \sqrt{5}}{2}=\left[\begin{array}{c}\mathrm{n}=\frac{1+\sqrt{5}}{2} \text { rejected } \\ \mathrm{n}=\frac{1-\sqrt{5}}{2}\end{array}\right. \\ & \therefore 4 \mathrm{x}^2-4 \mathrm{x}=4 \\ & (2 \mathrm{x}-1)^2=5\end{aligned}$

Let \(\mathrm{S}=\left\{x: \cos ^{-1} x=\pi+\sin ^{-1} x+\sin ^{-1}(2 x+1)\right\}\). Then \(\sum_{x \in…

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