$\begin{aligned} & \cos ^{-1}(2 \mathrm{x})=\pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2} \\\\ & \text { Since } \cos ^{-1}(2 \mathrm{x}) \in[0, \pi] \\\\ & \text { R.H.S. } \geq \pi \\\\ & \pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2}=\pi \\\\ & \Rightarrow \cos ^{-1} \sqrt{1-\mathrm{x}^2}=0 \\\\ & \Rightarrow \sqrt{1-\mathrm{x}^2}=1 \\\\ & \Rightarrow \mathrm{x}=0 \\\\ & \text { but at } \mathrm{x}=0 \\\\ & \cos ^{-1}(2 \mathrm{x})=\cos ^{-1}(0)=\frac{\pi}{2} \end{aligned}$ $\therefore$ No solution possible for given equation. $\mathrm{x} \in \phi$