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.

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.
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.

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.
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

∵f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
∴f(x)=2cos(x+π/3)sin(x+π/3)-2√3cos²(x+π/3)＝sin[2(x+π/3)]-√3cos[2(x+π/3)]－√3
＝2sin[2(x+π/3)－π/3]－√3＝2sin﹙2x－π/3﹚－√3
∵m[f(x)+√3]+2=0 ...

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∵f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
∴f(x)=2cos(x+π/3)sin(x+π/3)-2√3cos²(x+π/3)＝sin[2(x+π/3)]-√3cos[2(x+π/3)]－√3
＝2sin[2(x+π/3)－π/3]－√3＝2sin﹙2x－π/3﹚－√3
∵m[f(x)+√3]+2=0 ∴2msin﹙2x－π/3﹚＝﹣2
∴sin﹙2x－π/3﹚＝﹣1/m
∵x∈[0,π/6] ∴2x－π/3∈[﹣π/3，0] ∴sin﹙2x－π/3﹚∈[﹣√3/2，0]
∵对任意x属于[0,π/6],使得m[f(x)+√3]+2=0恒成立
∴﹣1/m∈[﹣√3/2，0]
∴1/m∈[0，√3/2]
∴m∈[2√3/3，﹢∞﹚

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