1、
f(x)=sin(2x-π/3)+√3/2=0
sin(2x-π/3)=-√3/2
2x-π/3=2kπ-π/3,2x-π/3=2kπ-2π/3
所以零点是x=kπ,x=kπ-π/6
2、
-1≤sin(2x-π/3)≤1
所以最小值是-1+√3/2,最大值是1+√3/2