Решите:
3+2sin3xsinx=3cos2x

3 + 2sin3xsinx = 3cos2x2sin3xsinx = 3cos2x - 3cos(3x - x) - cos(3x + x) = 3cos2x - 3cos2x - cos4x = 3cos2x - 3 -cos4x = 2cos2x - 3-(2cos²2x - 1) = 2cos2x - 3-2cos²2x - 2cos2x + 1 + 3 = 0cos²2x + cos2x - 2 = 0Пусть t = cos2x, t ∈ [-1; 1].t² + t - 2 = 0t₁ + t₂ = -1t₁t₂ = -2t₁ = -2 - не подходит; t₂ = 1Обратная замена:cos2x = 12x = 2πn, n ∈ Z x = πn, n ∈ Z Ответ: x = πn, n ∈ Z.