[latex]2cos ^{2} x-11sin2x=12[/latex]

Решение 2cos²x - 11sin2x = 12 2cos²x - 22sinxcosx - 12cos²x - 12sin²x = 0 12sin²x + 22sinxcosx + 10cos²x = 0 делим на 2cos²x ≠ 0 6tg²x + 11tgx + 5 = 0 tgx = t 6t² + 11t + 5 = 0 D = 121 - 4*6*5 = 1 t₁ = (11 - 1)/12 t₁ = 5/12 t₂ = (11 + 1)/12 t₂ = 1 1)  tgx = 5/12 x₁ = arctg(5/12) + πk, k ∈ Z tgx = 1 x₂ = π/4 + πn, n ∈ Z