Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equaion 

${\cos }^{-1}\left(x\right)-2{\sin }^{-1}\left(x\right)={\cos }^{-1}\left(2x\right)$ is equal to :

$$\begin{gathered} {\cos }^{-1}x=2 {\sin }^{-1}x={\cos }^{-1}2x \\ \boxed{{\cos }^{-1}x+ {\sin }^{-1}x=\frac{\mathrm{\pi }}{2}} \\ \Rightarrow {\cos }^{-1}x-2\left(\frac{\pi }{2}-{\cos }^{-1}x\right)={\cos }^{-1}2x \\ \Rightarrow {\cos }^{-1}x-\pi +2{\cos }^{-1}x={\cos }^{-1}2x \\ 3 {\cos }^{-1}x=\pi +{\cos }^{-1}2x ....... \left(1\right) \\ \therefore \cos \left(3 {\cos }^{-1}x\right)=\cos \left(\pi +{\cos }^{-1}2x\right) \\ \Rightarrow 4x^3-3x=-2x \\ \Rightarrow 4x^3=x\Rightarrow x=0, \pm \frac{1}{2} \\ \mathrm{All} \mathrm{satisfy} \mathrm{the} \mathrm{original} \mathrm{equation} \\ \therefore \mathrm{The} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{solutions}= -\frac{1}{2}+0+\frac{1}{2}=0 \end{gathered}$$