Home
О нас
Products
Services
Регистрация
Войти
Поиск
alexclark169
@alexclark169
July 2022
1
19
Report
Решите неравенства: 1) 2sin²x-3√2sinx+2>0;
2) 4sin²x+5sinx+2cos²x>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
sedinalana
1) 2sin²x-3√2sinx+2>0
sinx=t
2t
²-3√2t+2>0
D=18-16=2
t1=(3
√2-√2)/4=√2/2
t2=(3
√2+√2)/4=√2
t<
√2/2⇒sinx<√2/2⇒x∈(3π/4+2πk;9π/4+2πk.k∈z)
t>
√2⇒sinx>√2 нет решения
2) 4sin²x+5sinx+2cos²x>0
4sin
²x+5sinx+2-2sin²x>0
2sin²x+5sinx+2>0
sinx=t
2t²+5t+2>0
D=25-16=9
t1=(-5-3)/4=-2
t2=(-5+3)/4=-1/2
t<-2⇒sinx<-2 нет решения
t>-1/2⇒sinx>-1/2⇒x∈(-π/6+2πk;7π/6+2πk,k∈z)
0 votes
Thanks 1
alexclark169
А почему в 1) "t<√2/2⇒sinx<√2/2⇒x∈(3π/4+2πk;9π/4+2πk.k∈z)", там разве не t>√2/2, sinx>√2/2?
×
Report "Решите неравенства: 1) 2sin²x-3√2sinx+2>0; 2) 4sin²x+5sinx+2cos²x>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
sinx=t
2t²-3√2t+2>0
D=18-16=2
t1=(3√2-√2)/4=√2/2
t2=(3√2+√2)/4=√2
t<√2/2⇒sinx<√2/2⇒x∈(3π/4+2πk;9π/4+2πk.k∈z)
t>√2⇒sinx>√2 нет решения
2) 4sin²x+5sinx+2cos²x>0
4sin²x+5sinx+2-2sin²x>0
2sin²x+5sinx+2>0
sinx=t
2t²+5t+2>0
D=25-16=9
t1=(-5-3)/4=-2
t2=(-5+3)/4=-1/2
t<-2⇒sinx<-2 нет решения
t>-1/2⇒sinx>-1/2⇒x∈(-π/6+2πk;7π/6+2πk,k∈z)