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

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(B) Given the equation: $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$.Let $\sin ^{-1} x = 	heta$,then $x = \sin 	heta$. Since $x \in [-1, 1]$,$	he

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(B) Given the equation: $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$.Let $\sin ^{-1} x = 	heta$,then $x = \sin 	heta$. Since $x \in [-1, 1]$,$	heta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.The equation becomes $\sin ^{-1}(1-x) = \frac{\pi}{2} + 2	heta$.Taking $\sin$ on both sides: $1-x = \sin(\frac{\pi}{2} + 2	heta) = \cos(2	heta)$.Using the identity $\cos(2	heta) = 1 - 2\sin^2 	heta$,we get $1-x = 1 - 2x^2$.This simplifies to $2x^2 - x = 0$,which means $x(2x-1) = 0$.Thus,$x = 0$ 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 $x = \frac{1}{2}$: $\sin^{-1}(1-\frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = \sin^{-1}(\frac{1}{2}) - 2(\frac{\pi}{6}) = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} 
eq \frac{\pi}{2}$.Therefore,the only solution is $x = 0$.

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

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