JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Inverse Trigonometric Functions

Using ${\sin }^{-1}x=\frac{\pi }{2}-{\cos }^{-1}x$, rewrite as ${\cos }^{-1}x-2(\frac{\pi }{2}-{\cos }^{-1}x)={\cos }^{-1}(2x)$, which simplifies to $3{\cos }^{-1}x-\pi ={\cos }^{-1}(2x)$. Taking cosine of both sides: $\cos (3{\cos }^{-1}x)=\cos (\pi +{\cos }^{-1}(2x))$. Using the triple angle formula $\cos (3\theta )=4{\cos }^3\theta -3\cos \theta$: $4x^3-3x=-2x$. This gives $4x^3-x=0$, so $x(4x^2-1)=0$, yielding $x=0,\pm \frac{1}{2}$. All three values satisfy the original equation. The sum is $0+\frac{1}{2}-\frac{1}{2}=0$.