2cos²2x + cos2x + cos6x = 1
2cos²2x + cos2x + 4cos³2x - 3cos2x = 1
4cos³2x + 2cos²2x - 2cos2x - 1 = 0
2cos²2x(2cosx + 1) - (2cos2x + 1) = 0
(2cos²2x - 1)(2cosx + 1) = 0
Значение каждой скобки равно нулю, решаем два уравнения:
2cos²2x = 1
cos²2x = 1/2
cos2x = - √2/2
2x = ±3π/4 + 2πn, n ∈ Z
x = ±3π/8 + πn, n ∈ Z
cos2x = √2/2
2x = ±π/4 + 2πn, n ∈ Z
x = ±π/8 + πn, n ∈ Z
2cosx = -1
cosx = -1/2
x = ±2π/3 + 2πn, n ∈ Z
Ответ: x = ±2π/3 + 2πn, n ∈ Z; x = ±π/8 + πn, n ∈ Z; x = ±3π/8 + πn, n ∈ Z.