f(x)=(1+cos2ωx)/2+√3sin2ωx
=sin(2ωx+π/6)+1/2
T=2π/(2ω)=π,ω=1,f(x)=sin(2x+π/6)+1/2
（1）f(2π/3)=sin(4π/3+π/6)+1/2=-1/2
（2）2x+π/6∈[2kπ-π/2,2kπ+π/2],k∈Z
解得x∈[kπ-π/3,kπ+π/6],k∈Z时单调递增,增区间为(0,π/6]和[2π/3,π)
减区间为[π/6,2π/3]