已知A属于[0,2π],且满足sin(2A+π/6)+sin(2A-π/6)+2cos^2A>=2(1)求角A的取值集合M(2)若函数f(x)=cos2x+4ksinx (k>0,x属于M)最大值为3/2,求实数k的值
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![已知A属于[0,2π],且满足sin(2A+π/6)+sin(2A-π/6)+2cos^2A>=2(1)求角A的取值集合M(2)若函数f(x)=cos2x+4ksinx (k>0,x属于M)最大值为3/2,求实数k的值](/uploads/image/z/10157859-27-9.jpg?t=%E5%B7%B2%E7%9F%A5A%E5%B1%9E%E4%BA%8E%5B0%2C2%CF%80%5D%2C%E4%B8%94%E6%BB%A1%E8%B6%B3sin%282A%2B%CF%80%2F6%29%2Bsin%282A-%CF%80%2F6%29%2B2cos%5E2A%3E%3D2%EF%BC%881%EF%BC%89%E6%B1%82%E8%A7%92A%E7%9A%84%E5%8F%96%E5%80%BC%E9%9B%86%E5%90%88M%EF%BC%882%EF%BC%89%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%29%3Dcos2x%2B4ksinx+%28k%3E0%2Cx%E5%B1%9E%E4%BA%8EM%29%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA3%2F2%2C%E6%B1%82%E5%AE%9E%E6%95%B0k%E7%9A%84%E5%80%BC)
已知A属于[0,2π],且满足sin(2A+π/6)+sin(2A-π/6)+2cos^2A>=2(1)求角A的取值集合M(2)若函数f(x)=cos2x+4ksinx (k>0,x属于M)最大值为3/2,求实数k的值
已知A属于[0,2π],且满足sin(2A+π/6)+sin(2A-π/6)+2cos^2A>=2
(1)求角A的取值集合M
(2)若函数f(x)=cos2x+4ksinx (k>0,x属于M)最大值为3/2,求实数k的值
已知A属于[0,2π],且满足sin(2A+π/6)+sin(2A-π/6)+2cos^2A>=2(1)求角A的取值集合M(2)若函数f(x)=cos2x+4ksinx (k>0,x属于M)最大值为3/2,求实数k的值
(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