sin^2 2x+sin2xsinx+cos2x=1 x在第一象限,求cos x/2的值

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sin^22x+sin2xsinx+cos2x=1x在第一象限,求cosx/2的值sin^22x+sin2xsinx+cos2x=1x在第一象限,求cosx/2的值sin^22x+sin2xsinx+

sin^2 2x+sin2xsinx+cos2x=1 x在第一象限,求cos x/2的值
sin^2 2x+sin2xsinx+cos2x=1 x在第一象限,求cos x/2的值

sin^2 2x+sin2xsinx+cos2x=1 x在第一象限,求cos x/2的值
sin²(2x)+sin(2x)sin(x)+cos(2x)=1
sin²(2x)+sin(2x)sin(x)+1-2sin²(x)=1
sin²(2x)+sin(2x)sin(x)-2sin²(x)=0
[sin(2x)+2sin(x)][sin(2x)-sin(x)]=0
sin(2x)+2sin(x)=0
1-2sin²(x)+2sin(x)=0
2sin²(x)-2sin(x)-1=0
sin(x)=(1±√3)/2
sin(x)=(1+√3)/2>1 舍去
sin(x)=(1-√3)/2

=4sin^2xcos^2x+2sin^2xcosx+1-2sin^2x-1=0
=2sin^2xcos^2x+2sin^2xcosx-sin^2x=0
=2cos^2x+2cosx-1=0
=2(cosx+1/2)^2-3/2=0
所以(cosx+1/2)^2=3/4
因为x在第一象限
所以cosx>0
cosx+1/2=根号3/2
cosx=根号3/2-1/2
cox x/2=2cox^2x-1
剩下的你自己算吧