已知函数f(x)=sin(2x+π/6)+2sin²x
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/06/18 03:41:08
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
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
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)
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)
已知函数f(x)=2√3sin²x-sin(2x-π/3)
已知函数f(x)=sin(2x+π/6)+sin(2x+π/6)+2cos²x
已知函数f(x)=sin(2x+π/3)
已知函数f(x)=2sin(π-x)cosx
已知函数f(x)=sin(π/2-x)+sinx
已知函数f(x)=sinx+sin(x+π/2) ,
已知函数f(x)=2cos2x+sin²x
已知函数f(x)=2sin(π-x)sin(π/2-x)
高中数学:已知函数f(x)=2sin(x+π/2).sin(x+7π/3)-
已知函数f(x)=2sinx*sin(π/2+x)-2sin^2x+1
已知函数f(x)=sin(π-x)sin(π2-x)+cos2x
已知函数f(x)=2根号3sin平方x-sin(2x-π/3)