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Kate89
@Kate89
July 2022
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cos2x=1-cos(п/2-х)
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uomad4artem
cos2x=1-cos(п/2-х) =>
cos2x=1-sin(х) //////1-cos2x=2 sin ^2( x) подставим
2 sin ^2( x)-sin(х)=0
sin(х)=0 или 2 sin ( x)-1=0=> sin ( x)=1/2
x=pi*n или x = pi/6+ 2pi*n
x= 5pi/6+ 2pi*n
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Answers & Comments
cos2x=1-sin(х) //////1-cos2x=2 sin ^2( x) подставим
2 sin ^2( x)-sin(х)=0
sin(х)=0 или 2 sin ( x)-1=0=> sin ( x)=1/2
x=pi*n или x = pi/6+ 2pi*n
x= 5pi/6+ 2pi*n