已知函数f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
问题描述：
已知函数f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
对任意x属于[0,π/6],使得m[f(x)+√3]+2=0恒成立,求实数m的取值范围.
答
f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
=4cos(x+π/3)[1/2sin(x+π/3)-√3/2cos(x+π/3)]
=4cos(x+π/3)[sin(x+π/3-π/3)]
=4[cosxcos(π/3)-sinxsin(π/3)]*sinx
=2(cosx-√3sinx)sinx
=2cosxsinx-2√3sin²x
=sin2x+√3cos2x-√3
=2sin(2x+π/3)-√3,在[0,π/6]上,f(x)∈[1-√3,2-√3]
m[f(x)+√3]+2=0恒成立,即[f(x)+√3]=-2/m恒成立,所以1