решите пожалуйста 6cos^2x-2sin2x=1. Created by Dasdas123123123. algebra-ru

решите пожалуйста 6cos^2x-2sin2x=1...

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6сos²x - 2sin2x = 1
6cos²x - 4sinxcosx = sin²x + cos²x
sin²x + 4sinxcosx - 6cos²x + cos²x = 0
sin²x + 4sinxcosx - 5cos²x = 0 |:cos²x
tg²x + 4tgx - 5 = 0
Пусть t = tgx.
t² + 4t - 5 = 0
t₁ + t₂ = -4
t₁t₂ = -5
t₁ = -5; t₂ = 1
Обратная замена:
tgx = 1
x = π/4 + πn, n ∈ Z
tgx = -5
x = arctg(-5) + πk, k ∈ Z
Ответ: x = π/4 + πn, n ∈ Z; arctg(-5) + πk, k ∈ Z.

6 votes Thanks 11

sedinalana 6cos^2x-2sin2x=1
6сos²x-4sinxcosx-sin²x-cos²=0
sin²x+4sinxcosx-5cos²x=0/cos²x
tg²x+4tgx-5=0
tgx=t
t²+4t-5=0
D=16+20=36>0
t1+t2=-4 U t1*t2=-5
t1=-5⇒tgx=-5⇒x=-arctg5+πk,k∈z
t2=1⇒tgx=1⇒x=π/4+πk,k∈z

3 votes Thanks 3

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