Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Given equation is:

${\cos }^{-1}\left(2x\right)-2{\cos }^{-1}\left(\sqrt{1-x^2}\right)=\pi ,x\in \left[-\frac{1}{2},\frac{1}{2}\right]$

It is possible if

${\cos }^{-1}\left(2x\right)=\mathrm{\pi }$    ....(1)

and  $2{\cos }^{-1}\left(\sqrt{1-x^2}\right)=0$    ....(2)

From equation (1), $x=-\frac{1}{2}$ which is not satisfying equation (2).

So, no such $x$ exists.