$\begin{aligned} & \cos ^{-1} x=\pi+\sin ^{-1} x+\sin ^{-1}(2 x+1) \\ & 2 \cos ^{-1} x-\sin ^{-1}(2 x+1)=\frac{3 \pi}{2} \\ & 2 \alpha-\beta=\frac{3 \pi}{2} \text { where } \cos ^{-1} x=\alpha, \sin ^{-1}(2 x+1)=\beta \\ & 2 \alpha=\frac{3 \pi}{2}+\beta \\ & \cos 2 \alpha=\sin \beta \\ & \begin{array}{l} 2 \cos ^2 \alpha-1=\sin \beta \\ 2 x^2-1=2 x+1 \\ x^2-x-1=0 \\ \Rightarrow x=\frac{1-\sqrt{5}}{2},\left\{x=\frac{1+\sqrt{5}}{2} \text { rejected }\right\} \\ \therefore 4 x^2-4 x=4 \\ (2 x-1)^2=5 \end{array} \end{aligned}$