Given,
cos−1x−2sin−1x=cos−12x
⇒cos−1x−2(2π−cos−1x)=cos−12x
⇒cos−1x−π+2cos−1x=cos−12x
⇒3cos−1x=π+cos−12x...(1)
Taking cos both side we get,
⇒cos(3cos−1x)=cos(π+cos−12x)
⇒cos(cos−1(4x3−3x))=−cos(cos−12x)
⇒4x3−3x=−2x
⇒4x3=x
⇒x=0,±21
All will satisfy the original equation,
So, sum of all the solution will be 0+21−21=0