Cos^2x+2*sinx*cosx-3*sin^2x=0 помогите решить пожалуйста

Cos²x + 2 sinx cosx - 3 sin²x = 0 | : cos²x 1 + 2 tgx - 3tg²x = 0 | * (-1) 3tg²x - 2tgx -1 = 0 Пусть tgx = t, тогда 3t² - 2t -1 = 0 D = b² - 4ac = 4 + 12 = 16 [latex] x_{1} = \frac{2+4}{6} [/latex] [latex] x_{1} = 1[/latex] [latex] x_{2} = \frac{2 - 4}{6} [/latex] [latex] x_{2} = \frac{-1}{3} [/latex] Значит 1) tgx = 1 X = [latex] \frac{ \pi }{4} [/latex] + [latex] \pi n[/latex] 2) tgx = - [latex] \frac{1}{3} [/latex] x = - arctgx [latex] \frac{1}{3} [/latex] + [latex] \pi n[/latex] n ∈ Z