f(x)=(sinx)^4+(cosx)^2
=[(sinx)^2]^2+(cosx)^2
=[(1-cos2x)/2]^2+(1+cos2x)/2
=[1-2cos2x+(cos2x)^2+2+2cos2x]/4
=[3+(cos2x)^2]/4
=[3+(1+cos4x)/2]/4
=(1/8)cos4x+7/8
所以周期T=2π/4=π/2.最大值1最小值3/4