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gfdgfdg
@gfdgfdg
July 2022
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ewgenijbark
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Пусть х/2=α, тогда
sinα+sin3α=√3 cosα
sinα+3sinα-4sin³α-√3cosα=0
4sinα(1-sin²α)-√3cosα=0
4sinα cos²α-√3cosα=0
cosα(4sinαcosα-√3)=0
cosα=0 4sinαcosα-√3=0
α=π/2 +πn, n∈Z 2sinαcosα=√3/2
x/2=π/2+πn,n∈Z sin2α=√3 /2
x=π+2πn, n∈Z 2α=π/3 + 2πk, k∈Z
x1=π x=π/3 + 2πk, k∈Z
x2=π/3 +2π=7π/3
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Answers & Comments
Verified answer
Пусть х/2=α, тогдаsinα+sin3α=√3 cosα
sinα+3sinα-4sin³α-√3cosα=0
4sinα(1-sin²α)-√3cosα=0
4sinα cos²α-√3cosα=0
cosα(4sinαcosα-√3)=0
cosα=0 4sinαcosα-√3=0
α=π/2 +πn, n∈Z 2sinαcosα=√3/2
x/2=π/2+πn,n∈Z sin2α=√3 /2
x=π+2πn, n∈Z 2α=π/3 + 2πk, k∈Z
x1=π x=π/3 + 2πk, k∈Z
x2=π/3 +2π=7π/3