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batalygina
@batalygina
July 2022
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8sin^2x+3sinx*cosx+cos^2x=3
Срочно, пожалуйста
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Dимасuk
Verified answer
8sin²x + 3sinxcosx + cos²x = 3
8sin²x - 3 + 3sinxcosx + cos²x = 0
5sin²x + 3sinxcosx + cos²x - 3cos²x = 0
5sin²x + 3sinxcosx - 2cos²x = 0
Разделим на cos²x.
5tg²x + 3tgx - 2 = 0
Пусть t = tgx.
5t² + 3t - 2 = 0
D = 9 + 2•5•4 = 49 = 7²
t1 = (-3 + 7)/10 = 4/10 = 2/5
t2 = (-3 - 7)/10 = -1
Обратная замена:
tgx = 2/5
x = arctg(2/5) + πn, n ∈ Z
tgx = -1
x = -π/4 + πn, n ∈ Z.
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Answers & Comments
Verified answer
8sin²x + 3sinxcosx + cos²x = 38sin²x - 3 + 3sinxcosx + cos²x = 0
5sin²x + 3sinxcosx + cos²x - 3cos²x = 0
5sin²x + 3sinxcosx - 2cos²x = 0
Разделим на cos²x.
5tg²x + 3tgx - 2 = 0
Пусть t = tgx.
5t² + 3t - 2 = 0
D = 9 + 2•5•4 = 49 = 7²
t1 = (-3 + 7)/10 = 4/10 = 2/5
t2 = (-3 - 7)/10 = -1
Обратная замена:
tgx = 2/5
x = arctg(2/5) + πn, n ∈ Z
tgx = -1
x = -π/4 + πn, n ∈ Z.