Solve equation (8\cos^3 x+1)^3=162\cos x-27

Sarah-Louise Prince

Sarah-Louise Prince

Answered question

2022-02-27

Solve equation
(8cos3x+1)3=162cosx27

Answer & Explanation

Kathryn Duggan

Kathryn Duggan

Beginner2022-02-28Added 7 answers

This approach is based off knowing that the answers are 2π9, 4π9, 8π9. Another equation whose solutions are exactly those values, is 2cos3x+1=0, so we'd look to factorize that out.
We know that 2cos3x+1=8cos3x6cosx+1, so
(8cos3x+1)=(2cos3x+1+6cosx). Henceforth, let w=2cos3x+1
We have (w+6cosx)3=162cosx27, or that w3+3w26cosx+3w(6cosx)2+216cos3x=162cosx27=27w
hence w3+18w2cosx+108wcos2x+27w=0
This factorizes into 0=w(w2+18wcosx+108cos2x+27)
=w[(w+9cosx)2+27cos2x+27]
It is clear that the other term is strictly positive, so has no solution for x (regardless of what w is), so we must have w=0. Hence, the solutions are x=2π9, 4π9, 8π9

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