Home
О нас
Products
Services
Регистрация
Войти
Поиск
sonyakonstantinova
@sonyakonstantinova
July 2022
1
13
Report
решите уравнение 6sin^2x=5sinxcosx-cos^2x
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms of service
You must agree before submitting.
Send
Answers & Comments
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
×
Report "решите уравнение 6sin^2x=5sinxcosx-cos^2x..."
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
О нас
Политика конфиденциальности
Правила и условия
Copyright
Контакты
Helpful Social
Get monthly updates
Submit
Copyright © 2025 SCHOLAR.TIPS - All rights reserved.
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