Sinx * cosx + 2sin^2x=cos^2x

Решение sinx*cosx + 2sin²x = cos²x sinx*cosx + sin²x - (cos²x - sin²x) = 0 sinx*cosx + sin²x - (1 - 2sin²x) = 0 sinx*cosx + 3sin²x - 1 = 0 sinx*cosx + 3sin²x - sin²x - cos²x = 0 2sin²x + sinx*cosx - cos²x = 0      делим на cos²x ≠ 0 2tg²x + tgx - 1 = 0 tgx = t 2t² + t - 1 = 0 D = 1 + 4*2*1 = 9 t₁ = (-1 - 3)/4 t₁ = - 1 t₂ = (-1 + 3)/4 t₂ = 1/2 1)  tgx = - 1 x₁ = - π/4 + πk, k ∈ Z 2) tgx = 1/2 x₂ = arctg(1/2) + πn, n ∈ Z