Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation ${\cos }^{-1}\left(x\right)-2{\sin }^{-1}\left(x\right)={\cos }^{-1}\left(2x\right)$ is equal to

Given,

${\cos }^{-1}x-2{\sin }^{-1}x={\cos }^{-1}2x$

$\Rightarrow {\cos }^{-1}x-2\left(\frac{\pi }{2}-{\cos }^{-1}x\right)={\cos }^{-1}2x$

$\Rightarrow {\cos }^{-1}x-\pi +2{\cos }^{-1}x={\cos }^{-1}2x$

$\Rightarrow 3{\cos }^{-1}x=\pi +{\cos }^{-1}2x ...\left(1\right)$

Taking $\cos$ both side we get,

$\Rightarrow \cos \left(3{\cos }^{-1}x\right)=\cos \left(\pi +{\cos }^{-1}2x\right)$

$\Rightarrow \cos \left({\cos }^{-1}\left(4x^3-3x\right)\right)=-\cos \left({\cos }^{-1}2x\right)$

$\Rightarrow 4x^3-3x=-2x$

$\Rightarrow 4x^3=x$

$\Rightarrow x=0,\pm \frac{1}{2}$

All will satisfy the original equation,

So, sum of all the solution will be $0+\frac{1}{2}-\frac{1}{2}=0$