f(x)=sin(2x-π/4)-2根号2sin^2x
=sin(2x-π/4)-2根号(1-cos2x)
该式中第一项的最小正周期为π,后一项最小正周期也为π
因此f（x）的最小正周期为π
参考:
f(x)=根号2sin2x/2-根号2cos2x/2-2根号2(1-cos2x)/2
=根号2sin2x/2-根号2cos2x/2-根号2+根号2cos2x
=根号2sin2x/2+根号2cos2x/2-根号2
=sin(2x+π/4)-根号2
周期T=2π/w=2π/2=π