9cos2x - 27cosx + 18 = 0
9(2cos^2x-1)-27cosx+18 = 0
18cos^2x - 27cosx + 9 = 0
2cos^2x - 3cosx + 1 = 0
(2cosx-1)(cosx-1) = 0
cosx = 1/2 or 1
Take it from there.
Find all values of x in the interval [0, 2π] that satisfy the equation. (Enter your answers as a comma-separated list.)
18 + 9 cos(2x) = 27 cos(x)
2 answers
x=0, pi/3, 5pi/3