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ritka007
@ritka007
August 2022
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Помогите решить)
Ответ 4){п/2+kп,k принадлежит Z}
6)(0;4]
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irinan2014
Verified answer
Решение дано на фото.
0 votes
Thanks 1
sedinalana
4
сcosx/sinx-cosx=0
cosx(1-sinx)/sinx=0
ctgx*(1-sinx)=0
[ctgx=0⇒x=π/2+πk,k∈z
[1-sinx=0⇒sinx=1⇒π/2+2πk,k∈z
Общее x=π/2+πk,k∈z
6
16^(1/x)≥2
2^(4/x)≥2
4/x≥1
4/x-1≥0
(4-x)/x≥0
x=4 x=0
_ + _
--------------(0)-------------------[4]---------------
x∈(0;4]
2 votes
Thanks 1
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Answers & Comments
Verified answer
Решение дано на фото.сcosx/sinx-cosx=0
cosx(1-sinx)/sinx=0
ctgx*(1-sinx)=0
[ctgx=0⇒x=π/2+πk,k∈z
[1-sinx=0⇒sinx=1⇒π/2+2πk,k∈z
Общее x=π/2+πk,k∈z
6
16^(1/x)≥2
2^(4/x)≥2
4/x≥1
4/x-1≥0
(4-x)/x≥0
x=4 x=0
_ + _
--------------(0)-------------------[4]---------------
x∈(0;4]