JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Inverse Trigonometric Functions

On $\left[-\frac{1}{2},\frac{1}{2}\right]$, $\sqrt{1-x^2}\ge \frac{\sqrt{3}}{2}$, so $2{\cos }^{-1}\sqrt{1-x^2}$ lies in $\left[0,\frac{\pi }{3}\right]$ and the equation ${\cos }^{-1}(2x)=\pi +2{\cos }^{-1}\sqrt{1-x^2}$ needs a left side of at least $\pi$. Since ${\cos }^{-1}(2x)\le \pi$, equality would require ${\cos }^{-1}(2x)=\pi$ (that is $x=-\frac{1}{2}$) and ${\cos }^{-1}\sqrt{1-x^2}=0$ (that is $x=0$) simultaneously, which is impossible. The solution set is empty, so the sum taken over it is $0$.