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Lestat97
@Lestat97
June 2022
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2 cos^2 x=1+2 sin^2 x
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lfif02062002
Cos^2x - 1/2sin2x + cosx = sinxsin2x= 2sinx*cosxcos^2x- 1/2*2sinx*cosx+cosx = sinxcos^2x - 1/2*2sinx*cosx+cosx - sinx = 0cos^2x-sinx*cosx+cosx-sinx=0cosx(cosx+1) - sinx(cosx+1)=0(cosx+1)*(cosx-sinx)=0
cosx+1=0 -> cosx= -1 -> x=pi+2pi*K
cosx-sinx=0 Делим уравнение на корень из 2sin(pi/4-x)=0pi/4-x=pi*n
x=pi/4-pi*n
4 votes
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