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Irinasharova
@Irinasharova
August 2022
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2sin^2x - 10cos2x = 9 sin2x + 10
помогите решить уравнение
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Dимасuk
Verified answer
2sin²x - 10cos²x + 10sin²x = 18sinxcosx + 10
2sin²x - 10cos²x + 10sin²x - 10 - 18sinxcosx = 0
2sin²x - 10cos²x - 10cos²x - 18sinxcosx = 0
2sin²x - 18sinxcosx - 20cos²x = 0
Разделим на 2cos²x.
tg²x - 9tgx - 10 = 0
Пусть t = tgx.
t² - 9t - 10 = 0
t1 + t2 = 9
t1•t2 = -10
t1 = 10
t2 = -1
Обратная замена:
tgx = -1
x = -π/4 + πn, n ∈ Z
tgx = 10
x = arctg10 + πn, n ∈ Z.
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Verified answer
2sin²x - 10cos²x + 10sin²x = 18sinxcosx + 102sin²x - 10cos²x + 10sin²x - 10 - 18sinxcosx = 0
2sin²x - 10cos²x - 10cos²x - 18sinxcosx = 0
2sin²x - 18sinxcosx - 20cos²x = 0
Разделим на 2cos²x.
tg²x - 9tgx - 10 = 0
Пусть t = tgx.
t² - 9t - 10 = 0
t1 + t2 = 9
t1•t2 = -10
t1 = 10
t2 = -1
Обратная замена:
tgx = -1
x = -π/4 + πn, n ∈ Z
tgx = 10
x = arctg10 + πn, n ∈ Z.