Sin^2x+2sincosx-3cos^2x=0
sin²x + 2sinxcosx - 3cos²x = 0 | : cos²x
tg²x +2tgx - 3 = 0
Пусть tgx = t
t² + 2t - 3 = 0
D = 16
t₁ = ( - 2 + 4)/2 = 1;
t₂ = ( - 2 - 4)/2 = - 3;
Обратная замена:
tgx = - 3
x₁ = - arctg3 + πk, k ∈ Z
tgx = 1
x₂ = π/4 + πk, k ∈ Z