Solve 2sin 2x - cos 2x = ( tg x + 3)/(tgx + 1). - вопрос №890771122231 от King20051 27.06.2021 20:46

Question

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

King20051 · Answer

ОДЗ: Cosx ≠ 0    ⇒ x ≠ π/2 + πn,n ∈ z
         tgx + 1 ≠ 0 ⇒ tgx ≠ - 1   ⇒ x ≠ - π/4 + πn

$\frac{2*2tgx}{1+ tg^{2}x }- \frac{1- tg^{2} x}{1+ tg^{2}x } = \frac{tgx+3}{tgx+1} \\\\\frac{4tgx-(1- tg^{2} x)}{1+ tg^{2}x }= \frac{tgx+3}{tgx+1} \\\\ \frac{ tg^{2}x+4tgx-1 }{1+ tg^{2}x }= \frac{tgx+3}{tgx+1}$

(tg²x + 4tgx - 1)(tgx + 1) = (1+tg²x)(tgx + 3)
tg³x + tg²x + 4tg²x + 4tgx - tgx - 1 = tgx + 3 + tg³x + 3tg²x
5tg²x + 3tgx - 1 - tgx - 3 - 3tg²x = 0
2tg²x + 2tgx - 4 = 0
tg²x + tgx - 2 = 0
tgx₁ = - 2   ⇒ x = - arctg2 + πn , n ∈ z
tgx₂ = 1     ⇒ x = π/4 + πn , n ∈ z