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

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Aicharya · Accepted Answer

We have, $\begin{aligned} & \sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x \ & \Rightarrow \sin ^{-1}(1-x)=\frac{\pi}{2}-\sin ^{-1} x-\sin ^{-1} x \ & =\frac{\pi}{2}-2 \sin ^{-1} x \ & {\left[\because \cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}ight]} \ & \Rightarrow \sin ^{-1}(1-x)=\sin ^{-1} 1-\sin ^{-1} 2 x \sqrt{1-x^2} \ & \Rightarrow \sin ^{-1}(1-x)=\sin ^{-1}\left(1 \sqrt{1-4 x^2\left(1-x^2ight)}-0ight) \ & \Rightarrow (1-x)=\sqrt{1-4 x^4+4 x^2}=\sqrt{\left(1-2 x^2ight)^2} \ & \Rightarrow (1-x)=1-2 x^2 \ & \Rightarrow 2 x^2-x=0 \ & \Rightarrow x(2 x-1)=0 \ & \Rightarrow x=0, \frac{1}{2} \ & 	herefore x \in\left\{0, \frac{1}{2}ight\} \end{aligned}$

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

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