Question

Find the sum of all possible values of $x$ satisfying $\arccos \left(\frac{2}{\pi} \arccos x\right)=$ $\arcsin \left(\frac{2}{\pi} \arcsin x\right)$

Answer 1

$ \arccos \left(\frac{2}{\pi} \arccos x\right)=\arcsin \left(\frac{2}{\pi} \arcsin x\right)$
$\cos ^{-1}\left(\frac{2}{\pi}\left(\frac{\pi}{2}-\sin ^{-1} x\right)\right)=\sin ^{-1}\left(\frac{2}{\pi} \sin ^{-1} x\right) $
$\cos ^{-1}\left(1-\frac{2}{\pi} \sin ^{-1} x\right)=\sin ^{-1}\left(\frac{2}{\pi} \sin ^{-1} x\right)$
Let $\frac{2}{\pi} \sin ^{-1} x=\alpha$ where $\alpha \in[0,1]$ think!
$\Rightarrow \cos ^{-1}(1-\alpha)=\sin ^{-1} \alpha $
$\Rightarrow \sin ^{-1} \sqrt{2 \alpha-\alpha^2}=\sin ^{-1} \alpha \Rightarrow \sqrt{2 \alpha-\alpha^2}=\alpha \Rightarrow 2 \alpha-\alpha^2=\alpha^2$
$\Rightarrow 2 \alpha=2 \alpha^2$
Hence $\alpha$ is either 0 or 1 .
If $\alpha=0$ then $x=0$
if $\alpha=1$ then $x=1$
hence sum of all possible value of $x$ is 1