Home
О нас
Products
Services
Регистрация
Войти
Поиск
оса4621
@оса4621
July 2022
1
15
Report
cos2x=sinx-cosx тригонометрическое уравнение
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
Verified answer
Cos²x-sin²x-(sinx-cosx)=0
(cosx-sinx)(cosx+sinx)+(cosx-sinx)=0
(cosx-sinx)(cosx+sinx+1)=0
cosx-sinx=0/cosx
1-tgx=0
tgx=1
x=π/4+πn,n∈z
cosx+sinx+1=0
2cos²(x/2)+2sin(x/2)cos(x/2)=0
2cos(x/2)*(cos(x/2)+sin(x/2)=0
cos(x/2)=0
x/2=π/2+πk
x=π+2πk,k∈z
cos(x/2)+sin(x/2)=0/cos(x/2)
1+tg(x/2)=0
tg(x/2)=-1
x/2=-π/4+πm
x=-π/2+2πm,m∈z
12 votes
Thanks 23
рекомендуемые вопросы
rarrrrrrrr
August 2022 | 0 Ответы
o chem dolzhny pozabotitsya v pervuyu ochered vzroslye pri organizacionnom vyvoze n
danilarsentev
August 2022 | 0 Ответы
est dva stanka na kotoryh vypuskayut odinakovye zapchasti odin proizvodit a zapcha
myachina8
August 2022 | 0 Ответы
najti po grafiku otnoshenie v3v1 v otvetah napisano 9 no nuzhno reshenie
ydpmn7cn6w
August 2022 | 0 Ответы
Choose the correct preposition: 1.I am fond (out,of,from) literature. 2.where ar...
millermilena658
August 2022 | 0 Ответы
opredelite kak sozdavalas i kto sozdaval arabskoe gosudarstvo v kracii
MrZooM222
August 2022 | 0 Ответы
ch ajtmanov v rasskaze krasnoe yabloko ispolzuet metod rasskaz v rasskaze opi
timobila47
August 2022 | 0 Ответы
kakovo bylo naznachenie kazhdoj iz chastej vizantijskogo hrama pomogite pozhalujsta
ivanyyaremkiv
August 2022 | 0 Ответы
moment. 6....
pozhalujsta8b98a56c0152a07b8f4cbcd89aa2f01e 97513
sarvinozwakirjanova
August 2022 | 0 Ответы
pomogite pozhalusto pzha519d7eb8246a08ab0df06cc59e9dedb 6631
×
Report "cos2x=sinx-cosx тригонометрическое уравнение..."
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 © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Verified answer
Cos²x-sin²x-(sinx-cosx)=0(cosx-sinx)(cosx+sinx)+(cosx-sinx)=0
(cosx-sinx)(cosx+sinx+1)=0
cosx-sinx=0/cosx
1-tgx=0
tgx=1
x=π/4+πn,n∈z
cosx+sinx+1=0
2cos²(x/2)+2sin(x/2)cos(x/2)=0
2cos(x/2)*(cos(x/2)+sin(x/2)=0
cos(x/2)=0
x/2=π/2+πk
x=π+2πk,k∈z
cos(x/2)+sin(x/2)=0/cos(x/2)
1+tg(x/2)=0
tg(x/2)=-1
x/2=-π/4+πm
x=-π/2+2πm,m∈z