If sin^{-1}(1 - x) - 2sin^{-1}x = pi/2,then x | vedclass

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(C) Given the equation: $\sin^{-1}(1 - x) - 2\sin^{-1}x = \frac{\pi}{2}$ Rearranging the terms,we get: $\sin^{-1}(1 - x) = \frac{\pi}{2} + 2\sin^{-1}x

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(C) Given the equation: $\sin^{-1}(1 - x) - 2\sin^{-1}x = \frac{\pi}{2}$ Rearranging the terms,we get: $\sin^{-1}(1 - x) = \frac{\pi}{2} + 2\sin^{-1}x$ Taking $\sin$ on both sides: $1 - x = \sin\left(\frac{\pi}{2} + 2\sin^{-1}xight)$ Using the identity $\sin(\frac{\pi}{2} + 	heta) = \cos 	heta$: $1 - x = \cos(2\sin^{-1}x)$ Using the identity $\cos(2	heta) = 1 - 2\sin^2	heta$: $1 - x = 1 - 2\sin^2(\sin^{-1}x)$ Simplifying: $1 - x = 1 - 2x^2$ $2x^2 - x = 0$ $x(2x - 1) = 0$ So,$x = 0$ or $x = 1/2$. Checking $x = 1/2$: $\sin^{-1}(1 - 1/2) - 2\sin^{-1}(1/2) = \sin^{-1}(1/2) - 2(\pi/6) = \pi/6 - \pi/3 = -\pi/6 
eq \pi/2$. Checking $x = 0$: $\sin^{-1}(1 - 0) - 2\sin^{-1}(0) = \sin^{-1}(1) - 0 = \pi/2$. Thus,the only solution is $x = 0$.

If $\sin^{-1}(1 - x) - 2\sin^{-1}x = \pi/2$,then $x$ equals:

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