If cos (2sin ^{ - 1}x) = frac{1}{9}, then x = | vedclass

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(C) Given equation: $\cos (2\sin ^{ - 1}x) = \frac{1}{9}$ Let $\sin ^{ - 1}x = 	heta,$ then $\sin 	heta = x.$ The equation becomes $\cos (2	heta) =

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$\frac{2}{3}, -\frac{2}{3}$

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(C) Given equation: $\cos (2\sin ^{ - 1}x) = \frac{1}{9}$ Let $\sin ^{ - 1}x = 	heta,$ then $\sin 	heta = x.$ The equation becomes $\cos (2	heta) = \frac{1}{9}.$ Using the identity $\cos (2	heta) = 1 - 2\sin ^2	heta,$ we have: $1 - 2\sin ^2	heta = \frac{1}{9}$ $1 - 2x^2 = \frac{1}{9}$ $2x^2 = 1 - \frac{1}{9} = \frac{8}{9}$ $x^2 = \frac{4}{9}$ $x = \pm \frac{2}{3}.$ Since the range of $\sin ^{ - 1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}],$ the value of $2\sin ^{ - 1}x$ lies in $[-\pi, \pi].$ The cosine function is defined for both positive and negative values in this range,so both $x = \frac{2}{3}$ and $x = -\frac{2}{3}$ are valid solutions.

If $\cos (2\sin ^{ - 1}x) = \frac{1}{9},$ then $x = $

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