f(x)=cos^4x-2sinxcosx-sin^4x
=(cos^2x+sin^2x)(cos^2x-sin^2x)-sin2x
=1*cos2x-sin2x
=cos2x-sin2x
=cos(2x+π/4)
则f(x)的最小正周期
T=2π/2=π
f(x)=cos^4x-2sinxcosx-sin^4x
=(cos^2x+sin^2x)(cos^2x-sin^2x)-sin2x
=1*cos2x-sin2x
=cos2x-sin2x
=cos(2x+π/4)
则f(x)的最小正周期
T=2π/2=π