f(x)=(1-cos2x)/2+√3/2*sin2x+1/2
=(√3/2)sin2x-(1/2)cos2x+1
=sin2xcosπ/6-cos2xsinπ/6+1
=sin(2x-π/6)+1
所以当sin(2x-π/6)=1有最大值
所以2x-π/6=2kπ+π/2
2x=2kπ+2π/3
所以f(x)最大值=1+1=2
此时x=kπ+π/3