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Aikatsb
@Aikatsb
July 2022
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Решить уравнение: sin^2(x)-cos^2(x)=cos(x/2)
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kirichekov
Verified answer
Sin²x-cos²x=cos(x/2)
-(cos²x-sin²x)=cos(x/2)
-cos2x-cos(x/2)=0
cosα+cosβ=2cos((α+β)/2) *cos((α-β)/2)
-(cos2x+cos(x/2))=0
2cos(5/4)x *cos(3/4)x=0
cos(5/4)x=0 или cos(3/4)x=0
1. cos(5/4)x=0
5/4x=π/2+πn, n∈Z
x₁=2π/5+4πn/5, n∈Z
2. cos(3/4)x=0
3/4x=π/2+πn, n∈Z
x₂=2π/3+4πn/3, n∈Z
1 votes
Thanks 1
Aikatsb
в последнем период разве не 4/3Pi*n?
kirichekov
спасибо, иправила
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Answers & Comments
Verified answer
Sin²x-cos²x=cos(x/2)-(cos²x-sin²x)=cos(x/2)
-cos2x-cos(x/2)=0
cosα+cosβ=2cos((α+β)/2) *cos((α-β)/2)
-(cos2x+cos(x/2))=0
2cos(5/4)x *cos(3/4)x=0
cos(5/4)x=0 или cos(3/4)x=0
1. cos(5/4)x=0
5/4x=π/2+πn, n∈Z
x₁=2π/5+4πn/5, n∈Z
2. cos(3/4)x=0
3/4x=π/2+πn, n∈Z
x₂=2π/3+4πn/3, n∈Z