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Sezon
@Sezon
August 2022
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что делать с этим 2х? напишите подробно
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moboqe
Cos(2x-3pi/2)= -sin(2x)
-sin(2x)+2sqrt(3)cos^2(x)=0
-2sin(x)cos(x)+2sqrt(3)cos^2(x)=0
2cos(x)(sqrt(3)cosx-sin(x))=0
2cos(x)=0
cos(x)=0
x=pi/2+pi*k, k∈Z
sqrt(3)cosx-sin(x)=0
sqrt(3)cosx=sin(x) |cos(x)≠0
sqrt(3)=tg(x)
x=pi/3+pi*n, n∈Z
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Answers & Comments
-sin(2x)+2sqrt(3)cos^2(x)=0
-2sin(x)cos(x)+2sqrt(3)cos^2(x)=0
2cos(x)(sqrt(3)cosx-sin(x))=0
2cos(x)=0
cos(x)=0
x=pi/2+pi*k, k∈Z
sqrt(3)cosx-sin(x)=0
sqrt(3)cosx=sin(x) |cos(x)≠0
sqrt(3)=tg(x)
x=pi/3+pi*n, n∈Z