Помогите плиз 9tg^4x=6tg^2x-1 [-3п/2;0]

9tg^4x=6tg^2x-1              [-3п/2;0] 9tg^4x - 6tg^2x + 1 = 0   tg ²x = z 9z² - 6z + 1 = 0 D = 36 - 4*9*1 = 0 z ₁ = 6/18 z₁ = 1/3 tg ²x = 1/3 1)  tgx = - √3 x = - π/6 + πk, k∈Z 2) tgx = 1/√3 x = π/6 + πn, n∈Z   не принадлежит промежутку  [-3п/2;0] Ответ: x = - π/6 + πk, k∈Z