如下己知函数f（x）＝3sin平方x＋2根号3sinxcosx＋cos平方x,x属于R.（1）求函数f（x）的最大值与单调递增区间;（2）求使f（x）大于不等于3成立的x集合.

如下己知函数f（x）＝3sin平方x＋2根号3sinxcosx＋cos平方x,x属于R.（1）求函数f（x）的最大值与单调递增区间;（2）求使f（x）大于不等于3成立的x集合.
如下己知函数f（x）＝3sin平方x＋2根号3sinxcosx＋cos平方x,x属于R.
（1）求函数f（x）的最大值与单调递增区间;
（2）求使f（x）大于不等于3成立的x集合.

如下己知函数f（x）＝3sin平方x＋2根号3sinxcosx＋cos平方x,x属于R.（1）求函数f（x）的最大值与单调递增区间;（2）求使f（x）大于不等于3成立的x集合.
f(x)=3sin²x+2√3sinxcosx+cos²x
=3(1-cos2x)/2+√3sin2x+(1+cos2x)/2
=√3sin2x-cos2x+2
=2[sin2x*cos(π/6)-cos2x*sin(π/6)]+2
=2sin(2x-π/6)+2
(1) 当2x-π/6=2kπ+π/2,k∈Z时,
有最大值 4
增区间 2kπ-π/2≤2x-π/6≤2kπ+π/2
2kπ-π/3≤2x≤2kπ+2π/3
kπ-π/6≤x≤kπ+π/3
增区间【 kπ-π/6,kπ+π/3】k∈Z
（2）f(x)≥3
2sin(2x-π/6)+2≥3
sin(2x-π/6)≥1/2
2kπ+π/6≤2x-π/6≤2kπ+5π/6
2kπ+π/3≤2x≤2kπ+π
kπ+π/6≤x≤kπ+π/2
使f（x）大于不等于3成立的x集合.{x| kπ+π/6≤x≤kπ+π/2,k∈Z}

f（x）＝3sin平方x＋2根号3sinxcosx＋cos平方x
=2sin^2x+√3sin2x+1
=√3sin2x-cos2x+2
=2sin(2x-π/6)+2
(1) fmax=4
减区间 2kπ+π/2<=2x-π/6<=2kπ+3π/2
kπ+π/3<=x<=kπ+5π/6
[kπ+π/3，kπ+5π/6] k∈Z

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f（x）＝3sin平方x＋2根号3sinxcosx＋cos平方x
=2sin^2x+√3sin2x+1
=√3sin2x-cos2x+2
=2sin(2x-π/6)+2
(1) fmax=4
减区间 2kπ+π/2<=2x-π/6<=2kπ+3π/2
kπ+π/3<=x<=kπ+5π/6
[kπ+π/3，kπ+5π/6] k∈Z
2. 2sin(2x-π/6)+2<=3
sin(2x-π/6)+2<=1/2
2kπ-7π/6<=2x-π/6<=2kπ+π/6
kπ-π/2<=x<=kπ+π/6
成立的x集合 [kπ-π/2，kπ+π/6] k∈Z

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f(x)=3sin²x+2√3sinxcosx+cos²x=2sin²x+√3sin2x+1=1-cos2x+√3sin2x+1
=2sin(2x-π/6)+2
当2x-π/6=2kπ+π/2时，f(x)取得最大值2+2=4；
由：2kπ-π/2≤2x-π/6≤2kπ+π/2即：kπ-π/6≤x≤kπ+π/3时，f(x)递增；
所以f(...

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f(x)=3sin²x+2√3sinxcosx+cos²x=2sin²x+√3sin2x+1=1-cos2x+√3sin2x+1
=2sin(2x-π/6)+2
当2x-π/6=2kπ+π/2时，f(x)取得最大值2+2=4；
由：2kπ-π/2≤2x-π/6≤2kπ+π/2即：kπ-π/6≤x≤kπ+π/3时，f(x)递增；
所以f(x)的增区间为：[kπ-π/6,kπ+π/3];
f(x)＞3,即：sin(2x-π/6)>½; 则：2kπ+π/6<2x-π/6<2kπ+5π/6;
kπ+π/63成立的x的集合为( kπ+π/6,kπ+π/2)

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(1)f(x)=3sin²x+2√3sinxcosx+cos²x
=(√3sinx)²+2√3sinxcosx+cos²x
=(√3sinx+cosx）²
=4sin²(x+п/6)
∴f(x)的最大值为4
函数单调增...

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(1)f(x)=3sin²x+2√3sinxcosx+cos²x
=(√3sinx)²+2√3sinxcosx+cos²x
=(√3sinx+cosx）²
=4sin²(x+п/6)
∴f(x)的最大值为4
函数单调增区间为：2kп-п/2 ∴函数的单调增区间为[2kп-2п/3,2kп+п/3]
(2)4sin²(x+п/6)≥3
sin²(x+п/6)≥3/4
sin(x+п/6)≥√3/2或者sin(x+п/6)≤-√3/2
2Kп+п/3≤x+п/6≤2Kп+4п/3或者2Kп-4п/3≤x+п/6≤2Kп-п/3
所以满足条件的集合有[2kп+п/6,2kп+7п/6]∪[2kп-3п/2,2kп-п/2]

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3sin²x+2√3sinxcosx+cos²x
=2sin²x+2√3sinxcosx+1
=4sinx(1/2sinx+√3/2cosx)+1
=4sinx(cosπ/3sinx+sinπ/3cosx)+1
=4sinxsin(x+π/3)+1
=-2[cos(2x+π/3)-cos(π/3)+1
=-2cos(2x...

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3sin²x+2√3sinxcosx+cos²x
=2sin²x+2√3sinxcosx+1
=4sinx(1/2sinx+√3/2cosx)+1
=4sinx(cosπ/3sinx+sinπ/3cosx)+1
=4sinxsin(x+π/3)+1
=-2[cos(2x+π/3)-cos(π/3)+1
=-2cos(2x+π/3)+2
(1)函数最大值4，单调递增区间2x+π/3=[2nπ,（2n+1）π）
x=[nπ-π/6,nπ+π/3]
（2）求使f（x）大于不等于3成立的x集合.
f（x）=-2cos(2x+π/3)+2＞3
cos(2x+π/3)＜-1/2
2π/3+2nπ＜2x+π/3＜4π/3+2nπ
π/6+nπ＜x＜π/2+nπ

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