2cos2x = 4sin(π/2 + x) + 1
2cos2x = 4cosx + 1
4cos²x - 2 = 4cosx + 1
4cos²x - 4cosx - 3 = 0
4cos²x + 2cosx - 6cosx - 3 = 0
2cosx(2cosx + 1) - 3(2cosx + 1) = 0
(2cosx + 1)(2cosx - 3) = 0
1) 2cosx + 1 = 0
2cosx = -1
cosx = -1/2
x = ±2π/3 + 2πn, n ∈ Z
2) 2cosx - 3 = 0
2cosx = 3
cosx = 3/2 - нет корней, т.к. cosA ∈ [-1; 1]
Ответ: x = ±2π/3 + 2πn, n ∈ Z.