РЕШИТЕ ПРОШУ!!!!!!!!!!!!!cos2x=sin(x+пи/2)СРОЧНОБУДУ ОЧЕНЬ БЛАГОДАРНА!(решение нужно)
cos^2x-sin^2x=cosx
cos^2x-sin^2x-cosx=0
cos^2x-(1-cos^2x)-cosx=0
cos^2x-1+cos^2x-cosx=0
2cos^2x-cosx-1=0
cosx=t
2t^2-t-1=0
D=1-4*2*(-1)=9
x1=(1+3)/4=1
x2=(1-3)/4=-1/2
cosx=1 или cosx=-1/2
cosx=1
x=2pi
cosx=-1/2
x=2pi/3
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Answers & Comments
cos^2x-sin^2x=cosx
cos^2x-sin^2x-cosx=0
cos^2x-(1-cos^2x)-cosx=0
cos^2x-1+cos^2x-cosx=0
2cos^2x-cosx-1=0
cosx=t
2t^2-t-1=0
D=1-4*2*(-1)=9
x1=(1+3)/4=1
x2=(1-3)/4=-1/2
cosx=1 или cosx=-1/2
cosx=1
x=2pi
cosx=-1/2
x=2pi/3