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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
人气:314 ℃ 时间:2019-08-21 08:31:02
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
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