Home
О нас
Products
Services
Регистрация
Войти
Поиск
Илья228nsk
@Илья228nsk
August 2022
1
5
Report
cos^4X + sin^3X=1
Помогите пожалуйста!!
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
(1-sin^2x)^2+sin^3x=1;
1-2*sin^2x+(sin^2x)^2+sin^3x=1;
sin^4x+sin^3x-2*sin^2x=0;
sin^2x(sin^2x+sinx-2)=0;
sin^2x=0;
x1=0;
sin^2x+sinx-2=0;
sinx=u;
u^2+u-2=0;
D=9; u1=1; u2=-2;
sinx=1; x2=pi/2;
sinx=-2, x - не подх по одз
Ответ: x1=0; x2=pi/2
1 votes
Thanks 2
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 "cos^4X + sin^3X=1 Помогите пожалуйста!!..."
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
(1-sin^2x)^2+sin^3x=1;1-2*sin^2x+(sin^2x)^2+sin^3x=1;
sin^4x+sin^3x-2*sin^2x=0;
sin^2x(sin^2x+sinx-2)=0;
sin^2x=0;
x1=0;
sin^2x+sinx-2=0;
sinx=u;
u^2+u-2=0;
D=9; u1=1; u2=-2;
sinx=1; x2=pi/2;
sinx=-2, x - не подх по одз
Ответ: x1=0; x2=pi/2