Home
О нас
Products
Services
Регистрация
Войти
Поиск
potasss
@potasss
July 2022
1
1
Report
помогите пожалуйста
4 sin^2 2x+3=4cos^2 x
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
hlopushinairina
4sin²2x+3=4cos²x;⇒4·4sin²x·cos²x+3-4cos²x=0;⇒
4cos²x(4sin²x-1)+3=0⇒4cos²x·(4sin²x-4+3)+3=0⇒
4cos²x(3-4cos²x)+3=0;⇒12cos²x-16cos⁴x+3=0;⇒cos²x=t;⇒t>0;t≤1;
16t²-12t-3=0;
t₁,₂=[6⁺₋√(36+4·16·3)]/32=(6⁺₋√228)/32=(6⁺₋15.1)/32;
t₁=(6+15.1)/32=0.659;⇒cos²x=0.659;⇒cosx=⁺₋√659=⁺₋0.812;
x₁=⁺₋arccos(0.812)+2kπ;k∈Z;
x₂=⁺₋arccos(-0.812)+2kπ;k∈Z;
t₂=(6-15.1)-32=-0.284⇒t<0⇒решения нет
6 votes
Thanks 5
More Questions From This User
See All
potasss
July 2022 | 0 Ответы
pomogiite 2 cos4x 2 cos2x 4 cos2 x 1
Answer
potasss
October 2021 | 0 Ответы
v pravilnoj chetyrehugolnoj piramide sabcd tochka o centr osnovaniya s vershina4ffc0b2aa4d3c1803c285b1cca879f87 4103
Answer
potasss
October 2021 | 0 Ответы
pryamougolnyj parallelepiped opisan okolo cilindra radius osnovaniya kotorogo ra6424da1dcc681bf4212441a494a3c34c 85869
Answer
potasss
April 2021 | 0 Ответы
2 cos4x + 2 cos2x = 4 sin^2 x -1...
Answer
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 "помогите пожалуйста 4 sin^2 2x+3=4cos^2 x..."
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
4cos²x(4sin²x-1)+3=0⇒4cos²x·(4sin²x-4+3)+3=0⇒
4cos²x(3-4cos²x)+3=0;⇒12cos²x-16cos⁴x+3=0;⇒cos²x=t;⇒t>0;t≤1;
16t²-12t-3=0;
t₁,₂=[6⁺₋√(36+4·16·3)]/32=(6⁺₋√228)/32=(6⁺₋15.1)/32;
t₁=(6+15.1)/32=0.659;⇒cos²x=0.659;⇒cosx=⁺₋√659=⁺₋0.812;
x₁=⁺₋arccos(0.812)+2kπ;k∈Z;
x₂=⁺₋arccos(-0.812)+2kπ;k∈Z;
t₂=(6-15.1)-32=-0.284⇒t<0⇒решения нет