Home
О нас
Products
Services
Регистрация
Войти
Поиск
Katrinshark99
@Katrinshark99
August 2022
1
5
Report
Решите пожалуйста:
1) √3 cos3x = sin3x
2) 2 sin( π - 3x) + cos( 2π - 3x) = 0
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
Dимасuk
Verified answer
√3cos3x = sin3x (cos3x ≠ 0)
√3 = tg3x
3x = π/3 + πn, n ∈ Z
x = π/9 + πn/3, n ∈ Z
2sin(π - 3x) + cos(2π - 3x) = 0
-2sin(-3x) + cos(3x - 2π) = 0
2sin3x + cos3x = 0
2sin3x = -cos3x
2tg3x = -1
tg3x = -1/2
3x = arctg(-1/2) + πn, n ∈ Z
x = (arctg(-1/2))/3 + πn/3, n ∈ Z.
1 votes
Thanks 0
×
Report "Решите пожалуйста: 1) √3 cos3x = sin3x 2) 2 sin( π - 3x) + cos( 2π - 3x) = 0..."
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
Verified answer
√3cos3x = sin3x (cos3x ≠ 0)√3 = tg3x
3x = π/3 + πn, n ∈ Z
x = π/9 + πn/3, n ∈ Z
2sin(π - 3x) + cos(2π - 3x) = 0
-2sin(-3x) + cos(3x - 2π) = 0
2sin3x + cos3x = 0
2sin3x = -cos3x
2tg3x = -1
tg3x = -1/2
3x = arctg(-1/2) + πn, n ∈ Z
x = (arctg(-1/2))/3 + πn/3, n ∈ Z.