The number of solutions of sin ^{-1} x+sin ^{ | vedclass

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(C) Given equation: $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$. Taking $\sin$ on both sides: $\sin(\sin ^{-1} x + \sin ^{-1}(1-x)) = \sin(\cos ^{

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$2$

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(C) Given equation: $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$. Taking $\sin$ on both sides: $\sin(\sin ^{-1} x + \sin ^{-1}(1-x)) = \sin(\cos ^{-1} x)$. Using $\sin(A+B) = \sin A \cos B + \cos A \sin B$: $x \sqrt{1-(1-x)^2} + (1-x) \sqrt{1-x^2} = \sqrt{1-x^2}$. $x \sqrt{1-(1-2x+x^2)} + (1-x) \sqrt{1-x^2} = \sqrt{1-x^2}$. $x \sqrt{2x-x^2} + (1-x) \sqrt{1-x^2} = \sqrt{1-x^2}$. $x \sqrt{x(2-x)} = \sqrt{1-x^2} - (1-x) \sqrt{1-x^2} = x \sqrt{1-x^2}$. Squaring both sides: $x^2(2x-x^2) = x^2(1-x^2)$. $2x^3 - x^4 = x^2 - x^4$. $2x^3 - x^2 = 0 \Rightarrow x^2(2x-1) = 0$. So, $x = 0$ or $x = 1/2$. Checking $x=0$: $\sin^{-1}(0) + \sin^{-1}(1) = 0 + \pi/2 = \pi/2$. $\cos^{-1}(0) = \pi/2$. (Valid) Checking $x=1/2$: $\sin^{-1}(1/2) + \sin^{-1}(1/2) = \pi/6 + \pi/6 = \pi/3$. $\cos^{-1}(1/2) = \pi/3$. (Valid) Thus, there are $2$ solutions.

The number of solutions of $\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x$ is

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