Considering the principal values of the inverse trigonometric...

Given,

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

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

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

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

Taking $cos$ both side we get,

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

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

$⇒ 4{x}^{3} - 3x = - 2x$

$⇒ 4{x}^{3} = x$

$⇒ x = 0, \pm \dfrac{1}{2}$

All will satisfy the original equation,

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