Пожалуйста помогите решить уравнение 3sin2х - 4cоsх + 3sinх - 2 = 0. Укажите корни ,принадледащие отрезку [ П/2 ; 3П/2 ]

3sin2х - 4cоsх + 3sinх - 2 = 0 6sinх·cоsх - 4cоsх + 3sinх - 2 = 0 (6sinх·cоsх + 3sinх) - (4cоsх + 2) = 0 3sinх·(2cоsх + 1) - 2·(2cоsх + 1) = 0 (2cоsх + 1)·(3sinх - 2) = 0 1) 2cоsх + 1 = 0 cоsх = -1/2 x₁ = 4π/3 + 2πn x₂ = -4π/3 + 2πn 2) 3sinх - 2 = 0 sinх = 2/3 x₃ = (-1)^k ·arcsin(2/3) + πk Исследуем х₁ = 4π/3 + 2πn n = 0 x₁ = 4π/3  x∈[π/2; 3π/2] n = 1 x₁ = 4π/3 + 2π  x∉[π/2; 3π/2] Исследуем x₂ = -4π/3 + 2πn n = 1 x₂ = -4π/3 + 2π = 2π/3  x∈[π/2; 3π/2] n = 2  x₂ = -4π/3 + 4π = 8π/3  x∉[π/2; 3π/2] Исследуем x₃ = (-1)^k ·arcsin(2/3) + πk arcsin(2/3) ≈ 42° n = 1  x₃ = -arcsin(2/3) + π ≈ 138° x∈[π/2; 3π/2] n = 2  x₃ = arcsin(2/3) + 2π ≈ 402° x∉[π/2; 3π/2] Ответ: в интервале x∈[π/2; 3π/2] уравнеие имеет три корня x₁ = 4π/3, x₂ = 2π/3, x₃ = -arcsin(2/3) + π