Home
О нас
Products
Services
Регистрация
Войти
Поиск
Igor171717
@Igor171717
July 2022
2
8
Report
Нужно решить эти уравнения с подробным объяснением, заранее спасибо ВАМ
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
1)
sin2x = 3sinx*cos²x ;
sin2x =(
3/2)*2sinx*cosx *cosx ; * * * 2sinx*cosx =sin2x * * *
sin2x =(3/2)*sin2x*cosx
;
2sin2x = 3sin2x*cosx ;
2sin2x -3sin2x*cosx = 0 ;
3sin2x( 2/3 - cosx) =0 ;
[ sin2x =0 ; [ 2x =πk , k ∈ Z
[ 2/3 -cosx =0 . [ cosx = 2/3.
[ x =(π/2)*k , k ∈ Z ;
[ x = ± arccos(2/3) +2πn , n∈ Z .
--------------
2)
sin4x= sin2x * * * sin4x =sin2*2x = 2sin2x*cos2x * * *
2sin2x*cos2x =sin2x ;
2sin2x*cos2x - sin2x =0 ;
2sin2x(cos2x - 1/2) =0
;
[ sin2x = 0 ; [ 2x =πk , k ∈ Z ;
[cos2x =1/2 . [ 2x =±π/3 +2πn , n∈Z.
[ x =(π/2)*k , k ∈ Z ;
[ x =±π/6 +πn , n∈Z.
--------------
3)
cos2x+cos²x =0 ;
* * * cos2x =cos²x - sin²x =cos²x -(1 -cos²x) =2cos²x -1⇒
cos²x =(1+cos2x)/2 * *
cos2x+(1+cos2x)/2 =0 ;
2cos2x +1 +cos2x =0 ;
3cos2x = -1 ;
cos2x = -(1/3) ;
2x =±(π -arccos(1/3)) +2πk , k ∈Z.
x =±0,5(
π -arccos(1/3)) +πk , k ∈Z.
--------------
4)
sin2x =cos²x ;
2sinx*cosx =
cos²x ;
2sinx*cosx -
cos²x =0 ;
cosx(2sinx -cosx) =0 ;
[ cosx =0 ; [ x =π/2 +πk , k∈Z; [x=π/2 +πk , k∈Z;
[2sinx - cosx =0 . [
2sinx = cosx . [ tgx = 1/2 .
[x=π/2 +πk , k∈Z;
[
x = arctg(1/2) +
πn , n∈Z .
2 votes
Thanks 1
Igor171717
А почему вы делите?
Igor171717
sin2x =(3/2)*2sinx*cosx *cosx ; * * * 2sinx*cosx =sin2x * * *
Igor171717
А не умножаетее
oganesbagoyan
левая часть уравнения sin2x , а правая часть 3sinx*cos²x =(3/2)*2sinx*cosx*cosx || 3 =(3/2)*2 ||
oganesbagoyan
и sin2x =sin(x+x) =sinx*cosx +cosx*sinx =2sinx*cosx
sedinalana
1
sin2x=3sinxcos²x
2sin2x-1,5sin2xcosx=0
sin2x*(1-1,5cosx)=0
sin2x=0⇒2x=πk,k∈z⇒x=πk/2,k∈z
1-1,5cosx=0
1,5cosx=1
cosx=2/3⇒x=+-arccos2/3+2πk,k∈z
2
sin4x=sin2x
sin4x-sin2x=0
2sinx*cos3x=0
sinx=0⇒x=πk,k∈z
cos3x=0⇒3x=π/2+πk,k∈z⇒x=π/6+πk/3,k∈z
3
cos2x+cos²x=0
2cos²x-1+cos²x=0
cos²x-1=0
(cosx-1)9cosx+1)=0
cosx=1⇒x=2πk,k∈z
cosx=-1⇒x=3π/2+2πk,k∈z
4
sin2x=cos²z
2sinxcosx-cos²x=0
cosx*(2sinx-cosx)=0
cosx=0⇒x=π/2+πk,k∈z
2sinx-cosx=0/cosx
2tgx-1=0
2tgx=1
tgx=1/2
x=arctg1/2+πk,k∈z
0 votes
Thanks 0
More Questions From This User
See All
Igor171717
August 2022 | 0 Ответы
ustic u rastenij pustyn mnogo ili malo
Answer
Igor171717
August 2022 | 0 Ответы
provodyashij puchok lista eto i est ego zhilka zaputlsya
Answer
Igor171717
August 2022 | 0 Ответы
dobryj den mne neobhodima pomosh s biologiej dayu mnogo ballov tolko za razve
Answer
Igor171717
August 2022 | 0 Ответы
u geliofitov mnogo ili malo ustic
Answer
Igor171717
August 2022 | 0 Ответы
gde primenyayutsya izobrazheniya sobirayushej linzy dgt2f i d
Answer
Igor171717
August 2022 | 0 Ответы
devochka vospitannaya pochemu 2 n
Answer
Igor171717
August 2022 | 0 Ответы
najti tochki peregiba funkcii f x ln x21
Answer
Igor171717
August 2022 | 0 Ответы
vot polisaharidy aldegidogruppy i ketogruppy ne imeyut a disaharidyi eshyoch3c
Answer
Igor171717
August 2022 | 0 Ответы
vse li uglevody soderzhat gruppu oh
Answer
Igor171717
August 2022 | 0 Ответы
8...
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 "Нужно решить эти уравнения с подробным объяснением, заранее спасибо ВАМ..."
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)sin2x = 3sinx*cos²x ;
sin2x =(3/2)*2sinx*cosx *cosx ; * * * 2sinx*cosx =sin2x * * *
sin2x =(3/2)*sin2x*cosx ;
2sin2x = 3sin2x*cosx ;
2sin2x -3sin2x*cosx = 0 ;
3sin2x( 2/3 - cosx) =0 ;
[ sin2x =0 ; [ 2x =πk , k ∈ Z
[ 2/3 -cosx =0 . [ cosx = 2/3.
[ x =(π/2)*k , k ∈ Z ;
[ x = ± arccos(2/3) +2πn , n∈ Z .
--------------
2)
sin4x= sin2x * * * sin4x =sin2*2x = 2sin2x*cos2x * * *
2sin2x*cos2x =sin2x ;
2sin2x*cos2x - sin2x =0 ;
2sin2x(cos2x - 1/2) =0 ;
[ sin2x = 0 ; [ 2x =πk , k ∈ Z ;
[cos2x =1/2 . [ 2x =±π/3 +2πn , n∈Z.
[ x =(π/2)*k , k ∈ Z ;
[ x =±π/6 +πn , n∈Z.
--------------
3)
cos2x+cos²x =0 ;
* * * cos2x =cos²x - sin²x =cos²x -(1 -cos²x) =2cos²x -1⇒cos²x =(1+cos2x)/2 * *
cos2x+(1+cos2x)/2 =0 ;
2cos2x +1 +cos2x =0 ;
3cos2x = -1 ;
cos2x = -(1/3) ;
2x =±(π -arccos(1/3)) +2πk , k ∈Z.
x =±0,5(π -arccos(1/3)) +πk , k ∈Z.
--------------
4)
sin2x =cos²x ;
2sinx*cosx = cos²x ;
2sinx*cosx - cos²x =0 ;
cosx(2sinx -cosx) =0 ;
[ cosx =0 ; [ x =π/2 +πk , k∈Z; [x=π/2 +πk , k∈Z;
[2sinx - cosx =0 . [ 2sinx = cosx . [ tgx = 1/2 .
[x=π/2 +πk , k∈Z;
[x = arctg(1/2) +πn , n∈Z .
sin2x=3sinxcos²x
2sin2x-1,5sin2xcosx=0
sin2x*(1-1,5cosx)=0
sin2x=0⇒2x=πk,k∈z⇒x=πk/2,k∈z
1-1,5cosx=0
1,5cosx=1
cosx=2/3⇒x=+-arccos2/3+2πk,k∈z
2
sin4x=sin2x
sin4x-sin2x=0
2sinx*cos3x=0
sinx=0⇒x=πk,k∈z
cos3x=0⇒3x=π/2+πk,k∈z⇒x=π/6+πk/3,k∈z
3
cos2x+cos²x=0
2cos²x-1+cos²x=0
cos²x-1=0
(cosx-1)9cosx+1)=0
cosx=1⇒x=2πk,k∈z
cosx=-1⇒x=3π/2+2πk,k∈z
4
sin2x=cos²z
2sinxcosx-cos²x=0
cosx*(2sinx-cosx)=0
cosx=0⇒x=π/2+πk,k∈z
2sinx-cosx=0/cosx
2tgx-1=0
2tgx=1
tgx=1/2
x=arctg1/2+πk,k∈z