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Eloinka
@Eloinka
July 2022
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Помогите решить!!!!!
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ẞßnkt
2sin(x/2)=1-1+2sin^2(x/2)
Sin^2(x/2)-sin(x/2)=0
Sin(x/2)(sin(x/2)-1)=0
Sin(x/2)=0
X/2=pi*n,nпринадлежит z
X=2pi*n
Sin(x/2)=1
X/2=pi/2+2pi*n
X=pi+4pi*n
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Answers & Comments
Sin^2(x/2)-sin(x/2)=0
Sin(x/2)(sin(x/2)-1)=0
Sin(x/2)=0
X/2=pi*n,nпринадлежит z
X=2pi*n
Sin(x/2)=1
X/2=pi/2+2pi*n
X=pi+4pi*n