Question

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

Answer 1

解：(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
当sinx=-(√3)/2时 f(x)max=1-2*[-(√3)/2]^2+4k*[-(√3)/2]=3/2
k=-(√3)/3>-(√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<(√3)/2 矛盾！
综上所述： ∴k=±1/2