Объяснение:
РешениеSin²x + 5sinxcosx + 2cos²x = - 1 Sin²x + 5sinxcosx + 2cos²x = - sin²x - cos²x
2sin²x + 5sinxcosx + 3cos²x = 0 делим на cos²x ≠ 02tg²x + 5tgx + 3 = 0
tgx = t
2t² + 5t + 3 = 0
D = 25 - 4*2*3 = 1
t₁ = (- 5 - 1)/4 = - 6/4 = - 3/2 = - 1,5
t₂ = (- 5 + 1)/4 = - 1
1) tgx = - 3/2
x₁ = - arctg(1,5) + πk, k ∈ Z
2) tgx = - 1
x₂ = - π/4 + πn, n ∈ Z
