fancinate
幼苗
共回答了19个问题采纳率:94.7% 举报
(Ⅰ)∵f(x)=
3sinωx+cos(ωx+
π
3)+cos(ωx−
π
3)−1
=
3sinωx+[1/2]cosωx-
3
2sinωx+[1/2]cosωx+
3
2sinωx-1
=2sin(ωx+[π/6])-1,
∴函数f(x)的最小正周期为[2π/ω]=π;
∴ω=2.
∴f(x)=2sin(2x+[π/6])-1.
(Ⅱ)依题意,将函数f(x)昀图象向右平移[π/6]个单位,
得到函数g(x)=2sin(2x-[π/3]+[π/6])-1=2sin(2x-[π/6])-1的图象,
函数g(x)的解析式g(x)=2sin(2x-[π/6])-1.
∵0≤x≤[π/2],∴−
π
6≤2x-
1年前
1