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Yana4660
@Yana4660
July 2022
2
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Найти число решений уравнения на данном промежутке + решение.
[tex] tg^{2} x+3 ctg^{2} x=4, x-\ \textgreater \ (- \pi ; \pi )[/tex]
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sedinalana
Verified answer
Tgx≠0
tg²x=a
a+3/a=4
a²-4a+3=0
a1+a2=4 U a1*a2=3
a1=1⇒tg²x=1⇒tgx=-1 U tgx=1
x1=-π/4+πn,n∈z
-π<-π/4+πn<π
-4<-1+4n<4
-3<4n<5
-3/4<n<5/4
n=0⇒x=-π/4
n=1⇒x=-π/4+π=3π/4
x2=π/4+πk,k∈z
-π<π/4+πk<π
-4<1+4k<4
-5<4k<3
-5/4<k<3/4
k=-1⇒x=π/4-π=-3π/4
k=0⇒x=π/4
a2=3⇒tg²x=3⇒tgx=-√3 U tgx=√3
x3=-π/3+πm,m∈z
-π<-π/3+πm<π
-3<-1+3m<3
-2<3m<4
-2/3<m<4/3
m=0⇒x=-π/3
m=1⇒x=-π/3+π=2π/3
x4=π/3+πl,l∈z
-π<π/3+πl<π
-3<1+3l<3
-4<3l<2
-4/3<l<2/3
l=-1⇒x=π/3-π=-2π/3
l=0⇒x=π/3
1 votes
Thanks 1
alek48
ctg²x=1
x=П/4+Пn, n∈Z
ctg²x=1/3
x= arcctg1/√3+Пn, n∈Z
Что-то я накосячил .-.
0 votes
Thanks 0
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Answers & Comments
Verified answer
Tgx≠0tg²x=a
a+3/a=4
a²-4a+3=0
a1+a2=4 U a1*a2=3
a1=1⇒tg²x=1⇒tgx=-1 U tgx=1
x1=-π/4+πn,n∈z
-π<-π/4+πn<π
-4<-1+4n<4
-3<4n<5
-3/4<n<5/4
n=0⇒x=-π/4
n=1⇒x=-π/4+π=3π/4
x2=π/4+πk,k∈z
-π<π/4+πk<π
-4<1+4k<4
-5<4k<3
-5/4<k<3/4
k=-1⇒x=π/4-π=-3π/4
k=0⇒x=π/4
a2=3⇒tg²x=3⇒tgx=-√3 U tgx=√3
x3=-π/3+πm,m∈z
-π<-π/3+πm<π
-3<-1+3m<3
-2<3m<4
-2/3<m<4/3
m=0⇒x=-π/3
m=1⇒x=-π/3+π=2π/3
x4=π/3+πl,l∈z
-π<π/3+πl<π
-3<1+3l<3
-4<3l<2
-4/3<l<2/3
l=-1⇒x=π/3-π=-2π/3
l=0⇒x=π/3
ctg²x=1
x=П/4+Пn, n∈Z
ctg²x=1/3
x= arcctg1/√3+Пn, n∈Z
Что-то я накосячил .-.