已知函数f(x)=sin(2x+兀/6)+sin(2x-兀/6)+2cos^2x(1)求函数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
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![已知函数f(x)=sin(2x+兀/6)+sin(2x-兀/6)+2cos^2x(1)求函数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](/uploads/image/z/3007071-63-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%282x%2B%E5%85%80%2F6%29%2Bsin%282x-%E5%85%80%2F6%29%2B2cos%5E2x%281%29%E6%B1%82%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%8F%8A%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%282%29%E6%B1%82%E4%BD%BFf%28x%29%3E%3D2%E7%9A%84x%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
已知函数f(x)=sin(2x+兀/6)+sin(2x-兀/6)+2cos^2x(1)求函数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
已知函数f(x)=sin(2x+兀/6)+sin(2x-兀/6)+2cos^2x
(1)求函数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
已知函数f(x)=sin(2x+兀/6)+sin(2x-兀/6)+2cos^2x(1)求函数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
f(x)=sin(2x+π/6)+sin(2x-π/6)+2cos^2x
=sin2xcosπ/6+cos2xsinπ/6+sin2xcosπ/6-cos2xsinπ/6+2cos^2x
=√3sin2x+2cos^2x
=√3sin2x+1+cos2x
=2sin(2x+π/6)+1
最大值为3,最小正周期为π
2sin(2x+π/6)+1≥2
sin(2x+π/6)≥1/2
2x+π/6∈[π/6+2kπ,5π/6+2kπ]
x ∈ [kπ,π/3+kπ] k∈Z