f(x)=2cos^2ωx+2√3sinωxcosωx+3 　　=cos2ωx+√3sin2ωx+4 　　=2sin(2ωx+π/6)+4 　　(1)因为T=2π/2ω=π所以ω=1 　　(2)f(x)=2sin(2x+π/6)+4 　　单调增为2kπ-π/2≤2x+π/6≤2kπ+π/2 　　2kπ-4π/6≤2x≤2kπ+2π/6 　　kπ-π/3≤x≤kπ+π/6(k∈Z) 　　单调减为2kπ+π/2≤2x+π/6≤2kπ+3π/2 　　2kπ+2π/6≤2x≤2kπ+8π/6 　　kπ+π/6≤x≤kπ+2π/3(k∈Z) 　　故函数f(x)的单调递增区间：[kπ-π/3,kπ+π/6](k∈Z) 　　同理可得：函数f(x)的单调递减区间：[kπ+π/6,kπ+2π/3](k∈Z)