已知函数fx=cos(2x-π/3)+2sin(x-4)cos(x-π/4),x∈R若对任意x∈[-π/12,π/2]

问题描述:

已知函数fx=cos(2x-π/3)+2sin(x-4)cos(x-π/4),x∈R若对任意x∈[-π/12,π/2]
都有fx≥a成立,求a的取值范围

f(x)=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4),=1/2cos2x+√3/2sin2x+sin(2x-π/2)=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=sin(2x-π/6)因,f(x)≥asin(2x-π/6)≥ax∈[-π/12,π/2]2x-π/6∈[-π/3,5π/6]a...