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

Решение(3cos2x+5cosx-1)/√(- ctgx) = 0{3cos2x+5cosx-1 = 0{√(- ctgx) ≠ 0, ctgx ≠ 0, x ≠ π/2+ πk, k ∈ Z3cos2x+5cosx-1 = 03*(2cos²x - 1) + 5cosx - 1 = 06cos²x + 5cosx - 4 = 0cosx = t, I t I ≤ 16t² + 5t - 4 = 0D = 25 + 4*6*4 = 121t₁ = (- 5 - 11)/12t₁ = - 16/12 = - 4/3, не удовлетворяет условию I t I ≤ 1t₂ = (- 5 + 11)/12t₂ = 1/2cosx = 1/2x = (+-) arccos(1/2) + 2πn, n ∈ Zx = (+-) *(π/3) + 2πn, n ∈ ZОтвет: x = (+-) *(π/3) + 2πn, n ∈ Z,