Question

(3cos2x+5cosx-1)/$sqrt{-ctgx}$ =0

Answer 1

Решение
(3cos2x+5cosx-1)/√(- ctgx) = 0

{3cos2x+5cosx-1 = 0
{√(- ctgx) ≠ 0, ctgx ≠ 0, x ≠ π/2+ πk, k ∈ Z

3cos2x+5cosx-1 = 0
3*(2cos²x - 1) + 5cosx - 1 = 0
6cos²x + 5cosx - 4 = 0
cosx = t, I t I ≤ 1
6t² + 5t - 4 = 0
D = 25 + 4*6*4 = 121
t₁ = (- 5 - 11)/12
t₁ = - 16/12 = - 4/3, не удовлетворяет условию I t I ≤ 1
t₂ = (- 5 + 11)/12
t₂ = 1/2
cosx = 1/2
x = (+-) arccos(1/2) + 2πn, n ∈ Z
x = (+-) *(π/3) + 2πn, n ∈ Z

Ответ: x = (+-) *(π/3) + 2πn, n ∈ Z,