f(x)=sinx4+cosx4+sinx2cosx2/2-sin2x

法一:y = [(sinx)^4 + (cosx)^4 + (sinx)^2(cosx)^2]/2 - sin(2x)

= [(sinx)^4 + (cosx)^4 + 2(sinx)^2(cosx)^2 - (sinx)^2(cosx)^2]/2 - sin(2x)

= [1 - (sinx)^2(cosx)^2]/2 - sin(2x)

= 1/2 - [sin(2x)]^2/8 - sin(2x)

= {4 - 8sin(2x) - [sin(2x)]^2}/8

= {20 - 16 - 8sin(2x) - [sin(2x)]^2}/8

= {20 - [4 + sin(2x)]^2}/8

-5/8 = {20 - [4 + 1]^2}/8 <= y <= {20 - [4 - 1]^2}/8 = 11/8

x = kPI + PI/4时, y达到最小值 -5/8；
x = kPI + 3PI/4时,y达到最大值 11/8，其中k为任意整数。

y = {20 - [4 + sin(2x)]^2}/8,
最小正周期为π.

法二：令a=sinx,b=cosx,则a^2+b^2=1
分子=a^4+b^4+a^2b^2=a^4+2a^2b^2+b^4-a^2b^2
=(a^2+b^2)^2-a^2b^2
=1-a^2b^2
=1-(sinxcosx)^2
=1-(1/4)(sin2x)^2
=[1+(1/2)sin2x][1-(1/2)sin2x]

分母=2[[1-(1/2)sin2x]

所以f(x)=(1/2)+(1/4)sin2x
所以T=2π/2=π