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(2014•南开区二模)已知函数f(x)=sin(2x+π6)+sin(2x−π6)+2cos2x.

来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/05/10 03:23:39
(2014•南开区二模)已知函数f(x)=sin(2x+
π
6
)+sin(2x−
π
6
)+2cos
(Ⅰ)∵sin(2x+
π
6)=sin2xcos
π
6+cos2xsin
π
6,
sin(2x−
π
6)=sin2xcos
π
6−cos2xsin
π
6,cos2x=
1
2(cos2x+1)
∴f(x)=sin(2x+
π
6)+sin(2x−
π
6)+2cos2x
=sin2xcos
π
6+cos2xsin
π
6+sin2xcos
π
6−cos2xsin
π
6+cos2x+1
=
3sin2x+cos2x+1=2sin(2x+
π
6)+1
可得f(x)的最小正周期T=

|ω|=

2=π.
令−
π
2+2kπ≤2x+
π
6≤
π
2+2kπ(k∈Z),解之得−
π
3+kπ≤x≤
π
6+kπ(k∈Z),
∴函数f(x)的递增区间是[−
π
3+kπ,
π
6+kπ],k∈Z.
(Ⅱ)由f(x)≥2,得2sin(2x+
π
6)+1≥2(k∈Z),即sin(2x+
π
6)≥
1
2,
根据正弦函数的图象,可得
π
6+2kπ≤2x+
π
6≤

6+2kπ(k∈Z),
解之得kπ≤x≤kπ+