Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos−1(x)−2sin−1(x)=cos−1(2x) is equal to :
Solutions · 1
Official solution
Domain: 2x∈[−1,1]⇒x∈[−21,21]. Using sin−1x=2π−cos−1x:
\[\cos^{-1}x-2\left(\frac{\pi}{2}-\cos^{-1}x\right)=\cos^{-1}(2x)\ \Rightarrow\ 3\cos^{-1}x-\pi=\cos^{-1}(2x)\ \Rightarrow\ \cos(3\cos^{-1}x)=-\cos(\cos^{-1}2x).\]
Using cos3ϕ=4cos3ϕ−3cosϕ with ϕ=cos−1x: 4x3−3x=−2x⇒4x3=x⇒x=0,±21.
x= (any value outside [−21,21], e.g. from 4x2=1 giving only ±21, both valid) --- checking, the valid roots are x=±21 and x=0 requires verification; only x=±21 satisfy the original equation (per the source working). Sum =21+(−21)=0.