,向量m//向量n
2sin(ωx+ π/3)cosωx-1/2[(√3+2f(x)]=0
f(x)=2sin(ωx+ π/3)cosωx-√3/2
=(sinωx+√3cosωx)cosωx-√3/2
=sinωxcosωx+√3cos^2ωx-√3/2
=1/2sin2ωx+√3/2(1+cos2ωx)-√3/2
=1/2sin2ωx+√3/2cos2ωx
=sin(2ωx+π/3)
T=π/2*2=π
2ω=2π/π,ω=1
函数f(x)解析式.f(x)=sin(2x+π/3)
2x+π/3在[2kπ+π/2,2kπ+3π/2]单调递减
x在[kπ+π/12,kπ+7π/12]单调递减
x在［-π/3,π/6］
2x+π/3在［-π/3,2π/3］
的最大值＝1 ,最小值＝－√3/2