f(x)=sinωx+2√3sin2ωx/2的最小正周期为2π/3
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  • (1)

    f(x)=sinωx+2√3sin²ωx/2

    =sinwx+√3(1-coswx)

    =sinwx-√3coswx+√3

    =2(1/2sinwx-√3/2coswx)+√3

    =2sin(wx-π/3)+√3

    ∵f(x)最小正周期为2π/3

    ∴2π/w=2π/3

    ∴w=3

    ∴f(x)=2sin(3x-π/3)+√3

    (2)

    f(x)的图像向左平移π/2个单位,

    得g(x)=2sin[3(x+π/2)-π/3]+√3的图像

    ∴g(x)=2sin(3x+7π/6)+√3

    g(x)≥2√3

    即2sin(3x+7π/6)+√3≥2√3

    ∴sin(3x+7π/6)≥√3/2

    ∴2kπ+π/3≤3x+7π/6≤2kπ+2π/3,k∈Z

    ∴2kπ-5π/6≤3x≤2kπ-π/2,k∈Z

    ∴2kπ/3-5π/18≤x≤2kπ/3-π/6,k∈Z

    ∴g(x)≥2√3的解集为

    {x|2kπ/3-5π/18≤x≤2kπ/3-π/6,k∈Z}