Home
О нас
Products
Services
Регистрация
Войти
Поиск
stiklin
@stiklin
August 2022
1
36
Report
249(1,3)
250(1,3)
Спасибо !!!!!
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
Task/26534690
--------------------
249. 1)
250(1,3)
2cos(x/2) = 1+cos x ;
2cos(x/2) =2cos² (x/2) ;
2cos(x/2)*(cos(x/2) -1) = 0 ;
а)
cos(x/2) =0 ;
x/2= π/2 +πk , k ∈
ℤ ;
x = π +2πk , k ∈ ℤ .
б)
cos(x/2) -1 = 0 ;
cos(x/2) = 1 ;
x/2= 2πn , n ∈ ℤ ;
x = 4πn , n ∈ ℤ .
x = 2πk , k ∈ ℤ .
ответ :
2πk ;
π +2πk , k ∈ ℤ .
----------------------
249. 3)
sin3x / sinx = 0 ;
sinx (3 - 4sin²x) / sinx = 0 ; * * *ОДЗ: sinx ≠ 0* * *
3 - 4sin²x = 0 ;
3 - 4*(1-cos2x)/2 =0 ;
3 - 2 +2cos2x =0 ;
cos2x = - 1/2 ;
2x = ±(π-π/3) +2πn , n∈Z ;
2x = ±2π/3 +2πn , n∈Z ;
x = ±π/3 + πn , n∈Z .
ответ :
±π/3 + πn , n∈Z
.
----------------------
250. 1)
2sin²x +sinx =1 ;
2sin²x +sinx -1=0 ; квадратное уравнение относительно t =sinx
sinx₁ =(-1 -3)/4= -1 ⇒ x₁ = -π/2 +πn , n∈ℤ;
sinx₂=(-1 +3)/4=1/2 ⇒ x₂ = (-1)^n *π/6 +πn , n∈ℤ.
ответ :
-π/2 +πn ; (-1)^n *π/6 +πn , n∈Z. n∈
ℤ
.
----------------------
250. 3)
4sin⁴x +cos4x =1 +12cos⁴x ;
* * * имеем : sin²α =(1-cos2α)/2 ; cos²α =(1+cos2α)/2 * * *
4*
(
(1-cos2x)/2
)
²+cos4x = 1 +12*
(
(1+cos2x)/2
)
² ;
(1- cos2x)² + cos4x=1 +3(1+cos2x)² ;
1 - 2cos2x +cos²2x +cos4x =1 +3(1 +2cos2x +cos²2x);
1 +cos4x
+cos²2x - 2cos2x =3cos²2x +6co2x +4 ;
* * *1 +cos4x= 2cos²2x * * *
2cos²2x
+cos²2x - 2cos2x =3cos²2x +6co2x +4 ;
8cos2x = - 4 ;
cos2x = -1/2 ; уже решен в
249. 3)
x = ±π/3 + πn , n∈Z .
ответ :
± π/3 + πn , n∈Z
.
-------------
Удачи !
0 votes
Thanks 2
More Questions From This User
See All
stiklin
August 2022 | 0 Ответы
boring?...
Answer
stiklin
August 2022 | 0 Ответы
pomogite pozhalujsta perevesti na kazahskij ty zanimaeshsya kakim nibud vidom s
Answer
stiklin
August 2022 | 0 Ответы
zadaniie vo vlozhenii reshite pozhalujsta tam uravnenie
Answer
stiklin
August 2022 | 0 Ответы
ch kakaya skorost u vtoroj mashiny esli rasstoyanie mezhdu nimi 210 km reshit p
Answer
stiklin
August 2022 | 0 Ответы
sdelajte s risunkom dano najti i reshenie zaranee spasibo vysoty parallelogr
Answer
stiklin
August 2022 | 0 Ответы
tex]...
Answer
stiklin
August 2022 | 0 Ответы
perevesti na kazahskij yazyk pomozhesh mne s shkolnym proektom davaj kakoj p
Answer
stiklin
August 2022 | 0 Ответы
tex]...
Answer
stiklin
August 2022 | 0 Ответы
nomer 244polnostyu i nomer 24624
Answer
stiklin
August 2022 | 0 Ответы
pomogite reshit 2473 24813 bolshoe spasibo
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 "249(1,3) 250(1,3) Спасибо !!!!!..."
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
Task/26534690--------------------
249. 1)
250(1,3)2cos(x/2) = 1+cos x ;
2cos(x/2) =2cos² (x/2) ;
2cos(x/2)*(cos(x/2) -1) = 0 ;
а)
cos(x/2) =0 ;
x/2= π/2 +πk , k ∈ ℤ ;
x = π +2πk , k ∈ ℤ .
б)
cos(x/2) -1 = 0 ;
cos(x/2) = 1 ;
x/2= 2πn , n ∈ ℤ ;
x = 4πn , n ∈ ℤ .
x = 2πk , k ∈ ℤ .
ответ : 2πk ; π +2πk , k ∈ ℤ .
----------------------
249. 3)
sin3x / sinx = 0 ;
sinx (3 - 4sin²x) / sinx = 0 ; * * *ОДЗ: sinx ≠ 0* * *
3 - 4sin²x = 0 ;
3 - 4*(1-cos2x)/2 =0 ;
3 - 2 +2cos2x =0 ;
cos2x = - 1/2 ;
2x = ±(π-π/3) +2πn , n∈Z ;
2x = ±2π/3 +2πn , n∈Z ;
x = ±π/3 + πn , n∈Z .
ответ : ±π/3 + πn , n∈Z .
----------------------
250. 1)
2sin²x +sinx =1 ;
2sin²x +sinx -1=0 ; квадратное уравнение относительно t =sinx
sinx₁ =(-1 -3)/4= -1 ⇒ x₁ = -π/2 +πn , n∈ℤ;
sinx₂=(-1 +3)/4=1/2 ⇒ x₂ = (-1)^n *π/6 +πn , n∈ℤ.
ответ : -π/2 +πn ; (-1)^n *π/6 +πn , n∈Z. n∈ℤ .
----------------------
250. 3)
4sin⁴x +cos4x =1 +12cos⁴x ;
* * * имеем : sin²α =(1-cos2α)/2 ; cos²α =(1+cos2α)/2 * * *
4*( (1-cos2x)/2 )²+cos4x = 1 +12*( (1+cos2x)/2 )² ;
(1- cos2x)² + cos4x=1 +3(1+cos2x)² ;
1 - 2cos2x +cos²2x +cos4x =1 +3(1 +2cos2x +cos²2x);
1 +cos4x +cos²2x - 2cos2x =3cos²2x +6co2x +4 ;
* * *1 +cos4x= 2cos²2x * * *
2cos²2x +cos²2x - 2cos2x =3cos²2x +6co2x +4 ;
8cos2x = - 4 ;
cos2x = -1/2 ; уже решен в 249. 3)
x = ±π/3 + πn , n∈Z .
ответ : ± π/3 + πn , n∈Z .
-------------
Удачи !