MQ39 --- Trigonometry

Using the formula cos 2x = 2 cos^2 x - 1, the equation can be re-written as
cos 2x + cos 4x + cos 6x + 1 = 0
Since cos 2x + cos 6x = 2 cos 4x cos 2x, the equation becomes
2 cos 4x cos 2x + cos 4x + 1 = 0
Let cos 2x = y, so cos 4x = 2y^2 - 1, the equation becomes
2y(2y^2 - 1) + (2y^2 - 1) + 1 = 0
4y^3 - 2y + 2y^2 - 1 + 1 = 0
4y^3 + 2y^2 - 2y = 0
2y(2y^2 + y - 1) = 0
2y(2y - 1)(y + 1) = 0
that is :
(1) cos 2x = 0, x = (4n +/- 1)π/4. x = π/4, 3π/4, 5π/4, .....
(2) cos 2x = 1/2, x = (6n +/- 1)π/6. x = π/6, 5π/6, 7π/6,.....
(3) cos 2x = - 1, x = (2n +/- 1)π/2. x = π/2, 3π/2, ......

Your ans can be simpler actually...