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