Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Domain requires $0\le x\le 1$. Using ${\sin }^{-1}u=\pi /2-{\cos }^{-1}u$ and $\cos \left(2{\cos }^{-1}x\right)=x-1$ gives $2x^2-1=x-1$, or $x(2x-1)=0$. Both $\mathrm{x}=0$ and $\mathrm{x}=1/2$ satisfy the original equation.