已知A∈[0,π],且满足sin(2A+π/6)+sin(2A-π/6)+2cos ²A≥2

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  • (1)sin(2A+π/6)+sin(2A-π/6)+2cos^2A>=2

    (sin2Acosπ/6+cos2Asinπ/6)+(sin2Acosπ/6-cos2Asinπ/6)+cos2A+1≥2

    (√3)/2*sin2A+(√3)/2sin2A+cos2A≥2-1

    (√3)*sin2A+cos2A≥1

    2sin(2A+π/6)≥1

    sin(2A+π/6)≥1/2

    又∵ A∈[0,2π] 2A∈[0,4π] 2A+π/6∈[π/6,π/6+4π]

    2A+π/6∈[π/6,5π/6] ∪ [π/6+2π,5π/6+2π]

    A∈[0,π/3] ∪ [π,π/3+π]

    ∴M=[0,π/3] ∪ [π,π/3+π]

    (2)f(x)=cos2x+4ksinx=1-2(sinx)^2+4ksinx

    =-2(sinx-k)^2+2k^2+1

    ∵x=M∈[0,π/3] ∪ [π,π/3+π] sinx∈[ -(√3)/2,(√3)/2 ]

    (1)若k∈[ -(√3)/2,(√3)/2 ]

    当sinx=k时 f(x)max=2k^2+1=3/2

    k=±1/2=sinx∈[ -(√3)/2,(√3)/2 ]

    ∴k=±1/2

    (2)若k-(√3)/2 矛盾!

    (3)若k>(√3)/2

    当sinx=(√3)/2时 f(x)max=1-2*[(√3)/2]^2+4k*[(√3)/2]=3/2

    k=(√3)/3