题目

已知函数f(x)=sin(2x+π/6)+2sin²x
1.求函数f(x)的最小周期；2.求函数f(x)的最大值及取得最大值时x的取值集合；3.求函数f(x)的单调递增区间

提问时间：2021-01-12

答案

1.
f(x)=sin(2x+π/6)+2sin²x
=sin(2x+π/6)+1-cos(2x)
=sin(2x)cos(π/6)+cos(2x)sin(π/6) -cos(2x) +1
=sin(2x)cos(π/6)+(1/2)cos(2x) -cos(2x)+1
=sin(2x)cos(π/6)-(1/2)cos(2x) +1
=sin(2x)cos(π/6)-cos(2x)sin(π/6) +1
=sin(2x-π/6) +1
最小正周期=2π/2=π
2.
当sin(2x-π/6)=1时,f(x)有最大值[f(x)]max=1+1=2,此时2x-π/6=2kπ+π/2 (k∈Z)
x=kπ+π/3 (k∈Z)
当sin(2x-π/6)=-1时,f(x)有最小值[f(x)]min=-1+1=0,此时2x-π/6=2kπ-π/2 (k∈Z)
x=kπ-π/6 (k∈Z)
3.
2kπ-π/2≤2x-π/6≤2kπ+π/2 (k∈Z)时,函数单调递增
kπ-π/6≤x≤kπ+π/3 (k∈Z)
函数的单调递增区间为[kπ-π/6,kπ+π/3] (k∈Z)