Решите:

3+2sin3xsinx=3cos2x

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.