If sin ^{-1} x+sin ^{-1}(1-x)=cos ^{-1} x,the | vedclass

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(C) Given equation: $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$ Since $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$,we have: $\sin ^{-1}(1-x) = \f

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$\{0, \frac{1}{2}\}$

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(C) Given equation: $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$ Since $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$,we have: $\sin ^{-1}(1-x) = \frac{\pi}{2} - \sin ^{-1} x - \sin ^{-1} x = \frac{\pi}{2} - 2\sin ^{-1} x$ Taking $\sin$ on both sides: $1-x = \sin(\frac{\pi}{2} - 2\sin ^{-1} x) = \cos(2\sin ^{-1} x)$ Using the identity $\cos(2	heta) = 1 - 2\sin^2 	heta$,where $	heta = \sin ^{-1} x$: $1-x = 1 - 2(\sin(\sin ^{-1} x))^2$ $1-x = 1 - 2x^2$ $2x^2 - x = 0$ $x(2x - 1) = 0$ Thus,$x = 0$ or $x = \frac{1}{2}$. Checking the values: For $x=0$: $\sin ^{-1} 0 + \sin ^{-1} 1 = 0 + \frac{\pi}{2} = \frac{\pi}{2}$ and $\cos ^{-1} 0 = \frac{\pi}{2}$. (Valid) For $x=\frac{1}{2}$: $\sin ^{-1} \frac{1}{2} + \sin ^{-1} \frac{1}{2} = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$ and $\cos ^{-1} \frac{1}{2} = \frac{\pi}{3}$. (Valid) Therefore,$x \in \{0, \frac{1}{2}\}$.

If $\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x$,then $x \in$

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