f(x)=(√3sinωx+cosωx)cosωx-1/2=√3sinωx·cosωx+cos²ωx -1/2=(√3/2)sin2ωx+(1/2)cos2ωx=sin(2ωx+π/6)因为 周期为π，所以ω=1从而 f(x)=sin(2x+ π/6)令2kπ - π/2≤2x+π/6≤2kπ+π/2解得 kπ - π/3≤x≤kπ+π/6即增区间为 [kπ - π/3,kπ+π/6]，k是整数
