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sonyakonstantinova
@sonyakonstantinova
July 2022
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решите уравнение 6sin^2x=5sinxcosx-cos^2x
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nKrynka
Решение
6sin^2x=5sinxcosx-cos^2x
6sin^2x - 5sinxcosx + cos^2x = 0 делим на
²x ≠ 0
6tg
²x - 5tgx + 1 = 0
tgx = t
6t
² - 5t + 1 = 0
D = 25 - 4*6*1 = 1
t = (5-1)/12
t = 1/3
t = (5+1)/12
t = 1/2
1) tgx = 1/3
x
₁
= arctg(1/3) +
πk, k ∈ Z
2) tgx = 1/2
x
₂
= arctg(1/2) +
πn, n ∈Z
1 votes
Thanks 1
sonyakonstantinova
помогите еще с этим: 1)Найдите корни уравнения sin 3x + cos 3x = 0, принадлежащие отрезку [0, 6
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Answers & Comments
6sin^2x=5sinxcosx-cos^2x
6sin^2x - 5sinxcosx + cos^2x = 0 делим на ²x ≠ 0
6tg²x - 5tgx + 1 = 0
tgx = t
6t² - 5t + 1 = 0
D = 25 - 4*6*1 = 1
t = (5-1)/12
t = 1/3
t = (5+1)/12
t = 1/2
1) tgx = 1/3
x ₁= arctg(1/3) + πk, k ∈ Z
2) tgx = 1/2
x₂ = arctg(1/2) + πn, n ∈Z