Home
О нас
Products
Services
Регистрация
Войти
Поиск
Динара001
@Динара001
July 2022
1
7
Report
2sin^2x - cosx -1 = 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
AnonimusPro
Verified answer
2(1-cos^2x)-1-cosx=0;
-1-cosx+2(-cos^2x+1)=0;
-2cos^2x+1-cosx=0;
-2cos^2x-cosx=-1;
cosx=u;
-2u^2-u+1=0;
D=9;
u=1+-3/-4;
cosx=-(1+-3)/4;
x=arccos(-(1+-3)/4);
x=pi-arccos(1+-3/4);
x=pi-arccos(1-3/4) или pi-arccos(1+3/4);
x=pi или x=pi/3
2 votes
Thanks 5
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 "2sin^2x - cosx -1 = 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 © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Verified answer
2(1-cos^2x)-1-cosx=0;-1-cosx+2(-cos^2x+1)=0;
-2cos^2x+1-cosx=0;
-2cos^2x-cosx=-1;
cosx=u;
-2u^2-u+1=0;
D=9;
u=1+-3/-4;
cosx=-(1+-3)/4;
x=arccos(-(1+-3)/4);
x=pi-arccos(1+-3/4);
x=pi-arccos(1-3/4) или pi-arccos(1+3/4);
x=pi или x=pi/3