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

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