Considering the principal values of the inver | vedclass

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

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(A) Given equation: $\cos ^{-1}(x) - 2 \sin ^{-1}(x) = \cos ^{-1}(2x)$Using the identity $\sin ^{-1}(x) = \frac{\pi}{2} - \cos ^{-1}(x)$,we substitute:$\cos ^{-1}(x) - 2(\frac{\pi}{2} - \cos ^{-1}(x)) = \cos ^{-1}(2x)$Simplify the expression:$\cos ^{-1}(x) - \pi + 2 \cos ^{-1}(x) = \cos ^{-1}(2x)$$3 \cos ^{-1}(x) = \pi + \cos ^{-1}(2x)$Taking cosine on both sides:$\cos(3 \cos ^{-1}(x)) = \cos(\pi + \cos ^{-1}(2x))$Using the identity $\cos(3	heta) = 4\cos^3(	heta) - 3\cos(	heta)$ and $\cos(\pi + 	heta) = -\cos(	heta)$:$4x^3 - 3x = -2x$Rearrange the terms:$4x^3 - x = 0$$x(4x^2 - 1) = 0$Thus,the solutions are $x = 0$,$x = \frac{1}{2}$,and $x = -\frac{1}{2}$.Checking the solutions in the original equation:For $x = 0$: $\cos^{-1}(0) - 2\sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$ and $\cos^{-1}(0) = \frac{\pi}{2}$. (Valid)For $x = \frac{1}{2}$: $\cos^{-1}(\frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = \frac{\pi}{3} - 2(\frac{\pi}{6}) = 0$ and $\cos^{-1}(1) = 0$. (Valid)For $x = -\frac{1}{2}$: $\cos^{-1}(-\frac{1}{2}) - 2\sin^{-1}(-\frac{1}{2}) = \frac{2\pi}{3} - 2(-\frac{\pi}{6}) = \frac{2\pi}{3} + \frac{\pi}{3} = \pi$ and $\cos^{-1}(-1) = \pi$. (Valid)The sum of the solutions is $0 + \frac{1}{2} + (-\frac{1}{2}) = 0$.

Considering the principal values of the inverse trigonometric functions,the sum of all the solutions of the equation $\cos ^{-1}(x) - 2 \sin ^{-1}(x) = \cos ^{-1}(2x)$ is equal to.

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