3 + 2sin3xsinx = 3cos2x
2sin3xsinx = 3cos2x - 3
cos(3x - x) - cos(3x + x) = 3cos2x - 3
cos2x - cos4x = 3cos2x - 3
-cos4x = 2cos2x - 3
-(2cos²2x - 1) = 2cos2x - 3
-2cos²2x - 2cos2x + 1 + 3 = 0
cos²2x + cos2x - 2 = 0
Пусть t = cos2x, t ∈ [-1; 1].
t² + t - 2 = 0
t₁ + t₂ = -1
t₁t₂ = -2
t₁ = -2 - не подходит; t₂ = 1
Обратная замена:
cos2x = 1
2x = 2πn, n ∈ Z
x = πn, n ∈ Z
Ответ: x = πn, n ∈ Z.