Home
О нас
Products
Services
Регистрация
Войти
Поиск
FeniksHeat
@FeniksHeat
July 2022
1
4
Report
решите 4 и 5 номера пожалуйста завтра контрольная:(
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
oganesbagoyan
Verified answer
4.
а)
2cos²x -3cosx +1 =0 ;
Квадратное уравнение относительно cosx || можно и замену t=
cosx ||
[ cosx =1 ; cosx =1/2⇔ [
x =2πn ; x =±π/3 + 2πn , n∈Z
===========
б)
4sin²x +4cosx -1 =0 ;
4(1-cos²x) +
4cosx -1 =0 ;
4cos²x - 4cosx -3 =0 ; * * * cos²x - cosx -3/4 =0 * * *
cosx =( 2± √(2² -4*(-3) ) /4 =( 2±
4) /4
[ cosx = 3 /2 ; cosx = -1/2 ;
cosx = 3
/2 > 1 _не имеет решения
cosx = -1/2 ⇒
x = ±2π/3 + 2πn , n∈Z .
5.
√3sin4x +cos4x =0 ;
2(√3 /2 *sin4x +1/2*cos4x) =0 ;
2(√cos(π/6) *sin4x +sin(π/6)
*cos4x) =0 ;
2sin(4x+π/6) =0 ⇔sin(4x+π/6) =0⇒4x+π/6 =πn , n
∈Z ;
x = -
π/24 + (π/4)*n , n
∈Z .
1 votes
Thanks 2
More Questions From This User
See All
FeniksHeat
July 2022 | 0 Ответы
na risunke izobrazheny dve izotermy dlya dvuh gazov gazy mozhno schitat idealnymi
Answer
FeniksHeat
July 2022 | 0 Ответы
6)=-1...
Answer
FeniksHeat
July 2022 | 0 Ответы
vyruchajte rebyat zavtra kontrolnaya nado 3 5 nomera sdelat
Answer
FeniksHeat
July 2022 | 0 Ответы
reshite pliz 2 variant zadaniya 1 i 2 30 ballov dayu
Answer
FeniksHeat
July 2022 | 0 Ответы
1 kurs anglijskij yazyk sdelat 4 zadaniya s 1 po 3 nomer
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 "решите 4 и 5 номера пожалуйста завтра контрольная:(..."
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
4.а) 2cos²x -3cosx +1 =0 ;
Квадратное уравнение относительно cosx || можно и замену t=cosx ||
[ cosx =1 ; cosx =1/2⇔ [ x =2πn ; x =±π/3 + 2πn , n∈Z
===========
б) 4sin²x +4cosx -1 =0 ;
4(1-cos²x) + 4cosx -1 =0 ;
4cos²x - 4cosx -3 =0 ; * * * cos²x - cosx -3/4 =0 * * *
cosx =( 2± √(2² -4*(-3) ) /4 =( 2± 4) /4
[ cosx = 3 /2 ; cosx = -1/2 ;
cosx = 3 /2 > 1 _не имеет решения
cosx = -1/2 ⇒ x = ±2π/3 + 2πn , n∈Z .
5.
√3sin4x +cos4x =0 ;
2(√3 /2 *sin4x +1/2*cos4x) =0 ;
2(√cos(π/6) *sin4x +sin(π/6)*cos4x) =0 ;
2sin(4x+π/6) =0 ⇔sin(4x+π/6) =0⇒4x+π/6 =πn , n∈Z ;
x = - π/24 + (π/4)*n , n∈Z .