题目内容

已知函数f(x)=sin(2x+
π
6
)+sin(2x-
π
6
)+2cos2x(x∈R).
(1)求函数f(x)的最大值及此时自变量x的取值集合;
(2)求函数f(x)的单调递增区间;
(3)求使f(x)≥2的x的取值范围.
f(x)=sin2xcos
π
6
+cos2xsin
π
6
+sin2xcos
π
6
-cos2xsin
π
6
+1+cos2x=2sin2xcos
π
6
+cos2x+1=
3
sin2x+cos2x+1=2sin(2x+
π
6
)+1
(1)f(x)取得最大值3,此时2x+
π
6
=
π
2
+2kπ,即x=
π
6
+kπ,k∈Z
故x的取值集合为{x|x=
π
6
+kπ,k∈Z}
(2)由2x+
π
6
∈[-
π
2
+2kπ,
π
2
+2kπ],(k∈Z)得,x∈[-
π
3
+kπ,
π
6
+kπ],(k∈Z)
故函数f(x)的单调递增区间为[-
π
3
+kπ,
π
6
+kπ],(k∈Z)
(3)f(x)≥2?2sin(2x+
π
6
)+1≥2?sin(2x+
π
6
)≥
1
2
?
π
6
+2kπ≤2x+
π
6
6
+2kπ?kπ≤x≤
π
3
+kπ,(k∈Z)
故f(x)≥2的x的取值范围是[kπ,
π
3
+kπ],(k∈Z)
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