已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+2cos²x
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已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+2cos²x
(1)求f(x)的最大值及最小正周期;(2)求使f(x)≥2的x的取值范围.
(1)求f(x)的最大值及最小正周期;(2)求使f(x)≥2的x的取值范围.
(1)f(x)
=sin(2x+π/6)+sin(2x-π/6)+2cos²x
=sin2xcosπ/6+cos2xsinπ/6+sin2xcosπ/6-cos2xsinπ/6+2cos²x
=2sin2xcosπ/6+2cos²x
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
∵-1≤sin(2x+π/6)≤1
∴f(x)max=1×2+1=3
最小正周期T=2π/ω=2π/2=π
(2)
f(x)=2sin(2x+π/6)+1≥2
即:2sin(2x+π/6)≥1
sin(2x+π/6)≥1/2
∴π/6+2kπ≤2x+π/6≤5π/6+2kπ
即:{ x | kπ≤x≤π/3+kπ,k∈Z}
=sin(2x+π/6)+sin(2x-π/6)+2cos²x
=sin2xcosπ/6+cos2xsinπ/6+sin2xcosπ/6-cos2xsinπ/6+2cos²x
=2sin2xcosπ/6+2cos²x
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
∵-1≤sin(2x+π/6)≤1
∴f(x)max=1×2+1=3
最小正周期T=2π/ω=2π/2=π
(2)
f(x)=2sin(2x+π/6)+1≥2
即:2sin(2x+π/6)≥1
sin(2x+π/6)≥1/2
∴π/6+2kπ≤2x+π/6≤5π/6+2kπ
即:{ x | kπ≤x≤π/3+kπ,k∈Z}
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