sin(2x-π/3)>=0所以2kπ<=2x-π/3<=2kπ+π2kπ+π/3<=2x<=2kπ+4π/3kπ+π/6<=x<=kπ+2π/3tan(x+π/6)>0kπkπ-π/6所以定义域[kπ+π/6,kπ+π/3)
