<span>9tg^4x=6tg^2x-1 [-3п/2;0]
</span><span>9tg^4x - 6tg^2x + 1 = 0
tg</span>²x = z
9z² - 6z + 1 = 0
<span>D = 36 - 4*9*1 = 0
z</span>₁ = 6/18
z₁ = 1/3<span>
tg</span>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 <span> [-3п/2;0]
</span>Ответ: x = - π/6 + πk, k∈Z
A) Cos²x(2Cosx + √3) + (2Cosx + √3) = 0
(2Cosx +√3)(Cos²x +1) =0
2Cosx + √3 = 0 или Сos²x + 1 = 0
Cosx = -√3/2 Cos²x = -1
x = +-arcCos(-√3/2) + 2πk, k ∈Z нет решений.
х = +-5π/6 + 2πk , k ∈Z
б) [-2π; - π/2]
х = -7π/6; -5π/6