(2010•南昌模拟)已知函数f(x)=3sinxcosx+sin2x.
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(2010•南昌模拟)已知函数f(x)=
sinxcosx+sin
3 |
![(2010•南昌模拟)已知函数f(x)=3sinxcosx+sin2x.](/uploads/image/z/8260917-69-7.jpg?t=%EF%BC%882010%E2%80%A2%E5%8D%97%E6%98%8C%E6%A8%A1%E6%8B%9F%EF%BC%89%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D3sinxcosx%2Bsin2x%EF%BC%8E)
因为f(x)=
3sinxcosx+sin2x
=
3
2sin2x+
1-cos2x
2=
3
2sin2x-
1
2cos2x+
1
2
=sin2xcos
π
6cos2xsin
π
6+
1
2=sin(2x-
π
6)+
1
2,
(1)函数f(x)的最小正周期为T=
2π
2=π;
(2)当sin(2x-
π
6)=1时,f(x)取得最大值
3
2,
此时,2x-
π
6=2kπ+
π
2,k∈Z,
解得:x=kπ+
π
3,k∈Z,
∴f(x)的最大值为
3
2,取得最大值是x的集合为{x|x=kπ+
π
3,k∈Z};
(3)令2kπ-
π
2≤2x-
π
6≤2kπ+
π
2,k∈Z,
则2kπ-
π
3≤2x≤2kπ+
2π
3,k∈Z,
∴kπ-
π
6≤x≤kπ+
π
3,k∈Z,
∴f(x)的单调增区间为:[kπ-
π
6
3sinxcosx+sin2x
=
3
2sin2x+
1-cos2x
2=
3
2sin2x-
1
2cos2x+
1
2
=sin2xcos
π
6cos2xsin
π
6+
1
2=sin(2x-
π
6)+
1
2,
(1)函数f(x)的最小正周期为T=
2π
2=π;
(2)当sin(2x-
π
6)=1时,f(x)取得最大值
3
2,
此时,2x-
π
6=2kπ+
π
2,k∈Z,
解得:x=kπ+
π
3,k∈Z,
∴f(x)的最大值为
3
2,取得最大值是x的集合为{x|x=kπ+
π
3,k∈Z};
(3)令2kπ-
π
2≤2x-
π
6≤2kπ+
π
2,k∈Z,
则2kπ-
π
3≤2x≤2kπ+
2π
3,k∈Z,
∴kπ-
π
6≤x≤kπ+
π
3,k∈Z,
∴f(x)的单调增区间为:[kπ-
π
6
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