已知函数f(x)=-√3sinωxcosωx+cos²ωx,x∈R,ω>0⑴求函数f(x)的值域

f(x)= -√3sinωxcosωx+cos²ωx
=-(√3/2)sin(2ωx)+[1+cos(2ωx)]/2
=cos(2ωx)*cos(π/3)-sin(2ωx)*sin(π/3)+1/2
=cos(2ωx+π/3)+1/2
（1）函数的值域是[-1/2,3/2]
（2）最小正周期是T=2π/(2ω)=π/2
∴ ω=2
∴ f(x)=cos(4x+π/3)+1/2
先求所有的减区间
∴ 2kπ≤4x+π/3≤2kπ+π
∴ 2kπ-π/3≤4x≤2kπ+2π/3
∴ kπ/2-π/12≤x≤kπ/2+π/6
∵ x∈[0,π/2]
∴ 函数的单调减区间是[0,π/6]和[5π/12,π/2]