Solve sin ^{-1}(1-x)-2 sin ^{-1} x=frac{pi}{2 | vedclass

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(A) Given equation: $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$ Rearranging the terms: $-2 \sin ^{-1} x = \frac{\pi}{2} - \sin ^{-1}(1-x)$ Using th

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(A) Given equation: $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$ Rearranging the terms: $-2 \sin ^{-1} x = \frac{\pi}{2} - \sin ^{-1}(1-x)$ Using the identity $\frac{\pi}{2} - \sin ^{-1} 	heta = \cos ^{-1} 	heta$,we get: $-2 \sin ^{-1} x = \cos ^{-1}(1-x)$ Let $\sin ^{-1} x = y$,then $x = \sin y$. The equation becomes $-2y = \cos ^{-1}(1 - \sin y)$. Taking cosine on both sides: $\cos(-2y) = 1 - \sin y$ Since $\cos(-2y) = \cos(2y) = 1 - 2\sin^2 y$,we have: $1 - 2\sin^2 y = 1 - \sin y$ $2\sin^2 y - \sin y = 0$ $\sin y(2\sin y - 1) = 0$ This gives $\sin y = 0$ or $\sin y = \frac{1}{2}$. Since $x = \sin y$,possible values are $x = 0$ or $x = \frac{1}{2}$. Checking $x = 0$: $\sin^{-1}(1) - 2\sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. This is a solution. Checking $x = \frac{1}{2}$: $\sin^{-1}(\frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = -\sin^{-1}(\frac{1}{2}) = -\frac{\pi}{6} 
eq \frac{\pi}{2}$. Thus,the only solution is $x = 0$.

Solve $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2},$ then $x$ is equal to

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