(3cos2x+5cosx-1)/[latex] \sqrt{-ctgx} [/latex] =0

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,