f(x)=根号3sinxcosx-cos²x+1/2
=(√3/2)sin2x-[(1+cos2x)/2]+1/2
=(√3/2)sin2x-(1/2)cos2x
=sin(2x-π/6)
单调减区间：2kπ+π/2≤2x-π/6≤2kπ+3π/2,即kπ+π/3≤x≤kπ+5π/6