cos^2x-sin2x+根号3(2sinx*cosx)
=(1+cos2x)/2-sin2x+根号3sin2x
=[(根号3-1)sin2x+(1/2)cos2x]+1/2
=[(17-8根号3)/4]*sin(2x+w)+1/2
(此处tanw=1/[2(根号3-1)] )
则:最小正周期T=2pi/2=pi
由于:sin(2x+w)属于[-1,1]
则:[(17-8根号3)/4]sin(2x+w)
属于:[(8根号3-17)/4,(17-8根号3)/4]
则cos^2x-sin2x+根号3(2sinx*cosx)
=[(17-8根号3)/4]*sin(2x+w)+1/2
属于:[(8根号3-15)/4,(19-8根号3)/4]