Let S = { x : cos^{-1} x = pi + sin^{-1} x + | vedclass

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(A) Given the equation $\cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x + 1)$.Using the identity $\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$,we can rew

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$5$

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(A) Given the equation $\cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x + 1)$.Using the identity $\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$,we can rewrite $\cos^{-1} x$ as $\frac{\pi}{2} - \sin^{-1} x$.Substituting this into the equation: $\frac{\pi}{2} - \sin^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x + 1)$.Rearranging the terms: $-2 \sin^{-1} x - \sin^{-1}(2x + 1) = \frac{\pi}{2}$,or $2 \sin^{-1} x + \sin^{-1}(2x + 1) = -\frac{\pi}{2}$.Let $\sin^{-1} x = 	heta$,then $x = \sin 	heta$. The equation becomes $2	heta + \sin^{-1}(2x + 1) = -\frac{\pi}{2}$.$\sin^{-1}(2x + 1) = -\frac{\pi}{2} - 2	heta$.Taking $\sin$ on both sides: $2x + 1 = \sin(-\frac{\pi}{2} - 2	heta) = -\cos(2	heta)$.$2x + 1 = -(1 - 2\sin^2 	heta) = 2\sin^2 	heta - 1$.Since $x = \sin 	heta$,we have $2x + 1 = 2x^2 - 1$,which simplifies to $2x^2 - 2x - 2 = 0$,or $x^2 - x - 1 = 0$.The roots are $x = \frac{1 \pm \sqrt{5}}{2}$.Since the domain of $\sin^{-1} x$ is $[-1, 1]$ and $\sin^{-1}(2x + 1)$ requires $-1 \le 2x + 1 \le 1 \Rightarrow -2 \le 2x \le 0 \Rightarrow -1 \le x \le 0$,only $x = \frac{1 - \sqrt{5}}{2}$ is valid.For $x = \frac{1 - \sqrt{5}}{2}$,we have $2x - 1 = 1 - \sqrt{5} - 1 = -\sqrt{5}$.Thus,$(2x - 1)^2 = (-\sqrt{5})^2 = 5$.

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

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