f(x)＝cos^2x－√3 sinxcosx＋1
=（cos2x+1）/2-(√3/2)sin2x+1
=(1/2)cos2x-(√3/2)sin2x+3/2;
=cos(2x+π/3)+3/2;
∴单调递增时,-π+2kπ≤2x+π/3≤π+2kπ(k∈Z)
-4π/3+2kπ≤2x≤2π/3+2kπ;
-2π/3+kπ≤x≤π/3+kπ;
所以单调递增区间为[-2π/3+kπ,π/3+kπ](k∈Z)
如果本题有什么不明白可以追问,