f(x)=√3sin-cosx
=2[sinx*(√3/2)-cosx*(1/2)]
=2[sinxcos(π/6)-cosxsin(π/6)]
=2sin(x-π/6)
2sin(x-π/6)≥1
sin(x-π/6)≥1/2
2kπ+π/6≤x-π/6≤2kπ+5π/6
2kπ+π/3≤x≤2kπ+π
所以解集為{x|2kπ+π/3≤x≤2kπ+π,k∈Z}