On[0,2π) solve equation

Question

2cosxcos2x−2sinxsin2x=1

Answer 1

2 cos x cos 2x - 2 sin x sin 2x = 1

2(cos x cos 2x - sin x sin 2x) = 1 Sum for cos (x + 2x)

2 cos 3x = 1

cos 3x = 0.5

3x = arc cos 0.5

3x = π/3

x = π/9

Answer 2

2 cos (2cos^2-1) - 2 (sin)(2 sin cos) = 1

2 cos (2 cos^2-1) - 4 (sin^2 cos) = 1

4 cos^3- 2cos - 4 ( 1 - cos^2)(cos) = 1

4 cos^3 - 2cos - 4 cos + 4 cos^3 = 1

8 cos^3 - 6 cos - 1 = 0

Let U = cos^3

8 U^3 - 6U - 1 = 0

the approximate solutions are

U = -0.766

U = -0.174

U = 0.94

the inverse cosine of those are the solutions