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

Решение2cos²x - 11sin2x = 122cos²x - 22sinxcosx - 12cos²x - 12sin²x = 012sin²x + 22sinxcosx + 10cos²x = 0 делим на 2cos²x ≠ 06tg²x + 11tgx + 5 = 0tgx = t6t² + 11t + 5 = 0D = 121 - 4*6*5 = 1t₁ = (11 - 1)/12t₁ = 5/12t₂ = (11 + 1)/12t₂ = 11)  tgx = 5/12x₁ = arctg(5/12) + πk, k ∈ Ztgx = 1x₂ = π/4 + πn, n ∈ Z