Solve 2sin 2x - cos 2x = ( tg x + 3)/(tgx + 1).

ОДЗ: Cosx ≠ 0    ⇒  x ≠ π/2 + πn,n ∈ z         tgx + 1 ≠ 0  ⇒ tgx ≠ - 1   ⇒  x ≠ - π/4 + πn(tg²x + 4tgx - 1)(tgx + 1) = (1+tg²x)(tgx + 3)tg³x + tg²x + 4tg²x + 4tgx - tgx - 1 = tgx + 3 + tg³x + 3tg²x5tg²x + 3tgx - 1 - tgx - 3 - 3tg²x = 02tg²x + 2tgx - 4 = 0tg²x + tgx - 2 = 0tgx₁ = - 2   ⇒  x = - arctg2 + πn , n ∈ ztgx₂ = 1     ⇒  x = π/4 + πn , n ∈ z