作业帮已知A∈[0,π],且满足sin（2A+π/6）+sin（2A-π/6）+2cos ²A≥21.求角A的取值集合M 2.若函数f(x) =cos2x+4ksinx（k＞0,x∈M）的最大值是3/2,求实数k的值.
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/09 03:02:26
![已知A∈[0,π],且满足sin（2A+π/6）+sin（2A-π/6）+2cos ²A≥21.求角A的取值集合M 2.若函数f(x) =cos2x+4ksinx（k＞0,x∈M）的最大值是3/2,求实数k的值.](/uploads/image/z/8773092-36-2.jpg?t=%E5%B7%B2%E7%9F%A5A%E2%88%88%5B0%2C%CF%80%5D%2C%E4%B8%94%E6%BB%A1%E8%B6%B3sin%EF%BC%882A%2B%CF%80%2F6%EF%BC%89%2Bsin%EF%BC%882A-%CF%80%2F6%EF%BC%89%2B2cos+%26%23178%3BA%E2%89%A521.%E6%B1%82%E8%A7%92A%E7%9A%84%E5%8F%96%E5%80%BC%E9%9B%86%E5%90%88M+2.%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%29+%3Dcos2x%2B4ksinx%EF%BC%88k%EF%BC%9E0%2Cx%E2%88%88M%EF%BC%89%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E6%98%AF3%2F2%2C%E6%B1%82%E5%AE%9E%E6%95%B0k%E7%9A%84%E5%80%BC.)
已知A∈[0,π],且满足sin（2A+π/6）+sin（2A-π/6）+2cos ²A≥21.求角A的取值集合M 2.若函数f(x) =cos2x+4ksinx（k＞0,x∈M）的最大值是3/2,求实数k的值.
已知A∈[0,π],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∈M）的最大值是3/2,求实数k的值.
已知A∈[0,π],且满足sin（2A+π/6）+sin（2A-π/6）+2cos ²A≥21.求角A的取值集合M 2.若函数f(x) =cos2x+4ksinx（k＞0,x∈M）的最大值是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