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XXX1002
@XXX1002
July 2022
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Решите уравнение: cos^2x+sinx*sin(3pi/2+x)=1
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Annariel
cos^2(x)+sinx*sin(3pi/2+x)=1
cos^2(x)-sinx*cosx - 1 = 0
cos^2(x)-sinx*cosx- sin^2(x) - cos^2(x) = 0 |*(-1)
sinx*cosx+sin^2(x)=0
sinx(cosx+1)=0
sinx=0 и cosx+1=0
x=pi*n, ncZ cosx=-1 x=pi+2pi*k, kcZ
c - принадлежит
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Answers & Comments
cos^2(x)-sinx*cosx - 1 = 0
cos^2(x)-sinx*cosx- sin^2(x) - cos^2(x) = 0 |*(-1)
sinx*cosx+sin^2(x)=0
sinx(cosx+1)=0
sinx=0 и cosx+1=0
x=pi*n, ncZ cosx=-1 x=pi+2pi*k, kcZ
c - принадлежит