Home
О нас
Products
Services
Регистрация
Войти
Поиск
алисенью
@алисенью
July 2022
1
5
Report
решите уравнение
2cos^2 x - 3sin x=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
kirichekov
Verified answer
2cos²x-3sinx=0
2*(1-sin²x)-3sinx=0
2sin²x+3sinx-2=0
sinx=t, t∈[-1;1]
2t²+3t-2=0
D=3²-4*2*(-2)=25
t₁=(-3-5)/4, t₁=-2, -2∉[-1;1]
t₂=(-3+5)/4, t₂=1/2
sinx=1/2
x=(-1)^n*arcsin(1/2)+πn, n∈Z
x=(-1)^n*(π/6)+πn, n∈Z
15 votes
Thanks 22
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 "решите уравнение 2cos^2 x - 3sin x=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
2cos²x-3sinx=02*(1-sin²x)-3sinx=0
2sin²x+3sinx-2=0
sinx=t, t∈[-1;1]
2t²+3t-2=0
D=3²-4*2*(-2)=25
t₁=(-3-5)/4, t₁=-2, -2∉[-1;1]
t₂=(-3+5)/4, t₂=1/2
sinx=1/2
x=(-1)^n*arcsin(1/2)+πn, n∈Z
x=(-1)^n*(π/6)+πn, n∈Z