已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—2cos²x
已知函数f(x)=sin(2x+π/6)+sin(2x—π/6)—2cos²x
(1)求函数f(x)的值域及最小正周期.(2)求函数y=f(x)的单调增区间
(1)求函数f(x)的值域及最小正周期.(2)求函数y=f(x)的单调增区间
数学人气:470 ℃时间:2019-08-21 08:31:02
优质解答
f(x)=(√3/2)sin2x+(1/2)cos2x+(√3/2)sin2x-(1/2)cos2x-cos2x+1 =√3sin2x-cos2x+1 =2sin(2x-π/6)+1(1)sin(2x-π/6)∈【-1,1】所以,f(x)∈【-1,3】T=2π/2=π (2)-π/2+2kπ第一问是加一还是减一,再算遍不好意思,应该是-1f(x)=(√3/2)sin2x+(1/2)cos2x+(√3/2)sin2x-(1/2)cos2x-cos2x-1 =√3sin2x-cos2x-1 =2sin(2x-π/6)-1(1)sin(2x-π/6)∈【-1,1】所以,f(x)∈【-3,1】T=2π/2=π第2问?不好意思,也是符号写错了~~-π/2+2kπ<2x-π/6<π/2+2kπ-π/3+2kπ<2x<2π/3+2kπ-π/6+kπ
我来回答
类似推荐
- 已知函数f(x)=sin(2x+π6)+sin(2x-π6)+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.
- 已知函数f(x)=sin(2x+π/6)+sin(2x+π/6)+2cos²x
- 已知函数f(x)=sin(2x+π6)−cos(2x+π3)+2cos2x.(1)求f(π12)的值;(2)求f(x)的最大值及相应x的值.
- 已知函数f(x)=sin(2x+π6)−cos(2x+π3)+2cos2x.(1)求f(π12)的值;(2)求f(x)的最大值及相应x的值.
- 已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos²x-1,x∈R