f(x)
=2(cosx)^2+√3*sin2x [利用cos2x=2(cosx)^2-1化簡(jiǎn)]
=1+cos2x+√3*sin2x
=1+2[(1/2)*cos2x+(√3/2)*sin2x]
=1+2[sin(π/6)*cos2x+cos(π/6)*sin2x]
=1+2sin(2x+π/6)
最小正周期為 T=2π/ω=2π/2=π
單調(diào)遞增區(qū)間為：-π/2+2kπ≤2x+π/6≤π/2+2kπ (k∈整數(shù))
有：-π/3+kπ≤x≤π/6+kπ (k∈整數(shù))
原函數(shù)最小正周期為：π
單調(diào)遞增區(qū)間為：-π/3+kπ≤x≤π/6+kπ (k∈整數(shù))