Home
О нас
Products
Services
Регистрация
Войти
Поиск
Яноса
@Яноса
July 2022
1
18
Report
sin х-cosх>1............................................
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
avilov41
Sinx-cosx=√2(1/√2*sinx-1/√2cosx)=√2sin(x-π/4)
Решим неравенство √2sin(x-π/4)>1.
sin(x-π/4)>1/√2
π/4+2πn<x-π/4<3π/4+2πn,n∈z или
π/2+2πn<x<π+2πn,n∈z.
Ответ: π/2+2πn<x<π+2πn,n∈z.
0 votes
Thanks 1
×
Report "sin х-cosх>1..."
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
Решим неравенство √2sin(x-π/4)>1.
sin(x-π/4)>1/√2
π/4+2πn<x-π/4<3π/4+2πn,n∈z или
π/2+2πn<x<π+2πn,n∈z.
Ответ: π/2+2πn<x<π+2πn,n∈z.