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Loydsoiles
@Loydsoiles
July 2022
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99 БАЛЛОВ за полное и ПОДРОБНОЕ решение,подтягивайтесь ребята....................................
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dimanchyck2000
Verified answer
Заменяем log(3)2cosx на tПолучаем:2t² - 5t + 2 = 0t² - 2,5t + 1 = 0t1= 0,5t2 = 2log(3)2cosx = 0,5Log(3)2cosx = log(3)√32cosx = √3cosx = √3/2x = ± П/6 + 2 Пкlog(3)2cosx = log(3)92cosx = 9cosx = 4,5Решения нетНа промежутку П ; 5П/2 корни следующие:11П/6; 13П/6
5 votes
Thanks 3
sedinalana
Verified answer
ОДЗ
cosx>0
x∈(-π/2+2πn;π/2+2πn,n∈z)
log(3)(2cosx)=a
2a²-5a+2=0
D=25-16=9
a1=(5-3)/4=1/2
log(3)(2cosx)=1/2
2cosx=√3
cosx=√3/2
x=+-π/6+2πn
π≤-π/6+2πn≤5π/2
6≤-1+12n≤15
7≤12n≤16
7/12≤n≤4/3
n=1⇒x=-π/6+2π=11π/6
π≤π/6+2πn≤5π/2
6≤1+12n≤15
5≤12n≤14
5/12≤n≤7/6
n=1⇒x=π/6+2π=13π/6
a2=(5+3)/4=2
log(3)(2cosx)=2
2cosx=9
cosx=9/2>1нет решения
3 votes
Thanks 3
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Answers & Comments
Verified answer
Заменяем log(3)2cosx на tПолучаем:2t² - 5t + 2 = 0t² - 2,5t + 1 = 0t1= 0,5t2 = 2log(3)2cosx = 0,5Log(3)2cosx = log(3)√32cosx = √3cosx = √3/2x = ± П/6 + 2 Пкlog(3)2cosx = log(3)92cosx = 9cosx = 4,5Решения нетНа промежутку П ; 5П/2 корни следующие:11П/6; 13П/6Verified answer
ОДЗcosx>0
x∈(-π/2+2πn;π/2+2πn,n∈z)
log(3)(2cosx)=a
2a²-5a+2=0
D=25-16=9
a1=(5-3)/4=1/2
log(3)(2cosx)=1/2
2cosx=√3
cosx=√3/2
x=+-π/6+2πn
π≤-π/6+2πn≤5π/2
6≤-1+12n≤15
7≤12n≤16
7/12≤n≤4/3
n=1⇒x=-π/6+2π=11π/6
π≤π/6+2πn≤5π/2
6≤1+12n≤15
5≤12n≤14
5/12≤n≤7/6
n=1⇒x=π/6+2π=13π/6
a2=(5+3)/4=2
log(3)(2cosx)=2
2cosx=9
cosx=9/2>1нет решения