Answer:
See below in bold.
Step-by-step explanation:
cos 4x = 2cos^2 2x - 1 = 2 (2 cos^2 x - 1)^2 - 1
and cos 2x = 2 cos^2 x - 1 so we have:
2 ( 2 cos^2 x - 1)^2 - 1 - (2cos^2 x - 1) = 0
2 ( 2 cos^2 x - 1)^2 - 2 cos^2 x = 0
(2 cos^2 x - 1)^2 - cos^2 x = 0
Let c = cos^2 x, then:
(2c - 1)^2 - c = 0
4c^2 - 4c + 1 - c = 0
4c^2 - 5c + 1 = 0
c = 0.25, 1
cos^2 x = 0.25 gives cos x = +/- 0.5
and cos^2 x = 1 gives cos x = +/- 1.
So for x = +/- 1 , x = 0, π.
For cos x = +/- 0.5, x = π/3, 2π/3, 4π3,5π/3.
