Home
О нас
Products
Services
Регистрация
Войти
Поиск
Veronichka1998
@Veronichka1998
August 2022
2
14
Report
Помогите пожалуйста! решите тригонометрическое уравнение:
2cos²x-11sin2x=12
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
Лизаветк132
А ты в каком классе учишься???
0 votes
Thanks 0
sofiya3009
Не знаю правильно это или нет
Решение
2cos²x - 11sin2x = 12
2cos²x - 22sinxcosx - 12cos²x - 12sin²x = 0
12sin²x + 22sinxcosx + 10cos²x = 0 делим на 2cos²x ≠ 0
6tg²x + 11tgx + 5 = 0
tgx = t
6t² + 11t + 5 = 0
D = 121 - 4*6*5 = 1
t₁ = (11 - 1)/12
t₁ = 5/12
t₂ = (11 + 1)/12
t₂ = 1
1) tgx = 5/12
x₁ = arctg(5/12) + πk, k ∈ Z
tgx = 1
x₂ = π/4 + πn, n ∈ Z
0 votes
Thanks 0
×
Report "Помогите пожалуйста! решите тригонометрическое уравнение: 2cos²x-11sin2x=12..."
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
Решение
2cos²x - 11sin2x = 12
2cos²x - 22sinxcosx - 12cos²x - 12sin²x = 0
12sin²x + 22sinxcosx + 10cos²x = 0 делим на 2cos²x ≠ 0
6tg²x + 11tgx + 5 = 0
tgx = t
6t² + 11t + 5 = 0
D = 121 - 4*6*5 = 1
t₁ = (11 - 1)/12
t₁ = 5/12
t₂ = (11 + 1)/12
t₂ = 1
1) tgx = 5/12
x₁ = arctg(5/12) + πk, k ∈ Z
tgx = 1
x₂ = π/4 + πn, n ∈ Z