设函数f(x)=cosωx((√ 3)sinωx+cosωx),其中0＜ω＜2.求

f(x)=cosωx((√3)sinωx+cosωx)=(√3)sinωx*cosωx+(cosωx)^2=(√3/2)sin2ωx+(1/2)cosωx+1/2
=sin(2ωx+π/6)+1/2
f(x)周期为π,2π/2ω=π,ω=1
f(x)==sin(2x+π/6)+1/2
令2kπ-π/2≤2x+π/6≤2kπ+π/2,
kπ-π/3/≤x≤kπ+π/6,(k为整数）
f(x)在[-π/6,π/6]单增,[π/6,π/3]单减
在x=-π/6有最小值0,在x=π/6有最大值3/2
值域为[0,3/2]