f(x)=√3cos²x＋sinxcosx
=√3/2*(cos2x+1)+1/2(sin2x)
=1/2sin2x+√3/2cos2x+√3/2
=sin(2x+π/3)+√3/2
当2x+π/3=2kπ+π/2,k∈Z,取最大值1+√3/2；
当2x+π/3=2kπ-π/2,k∈Z,取最大值√3/2-1.