已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a(a∈R,a为常数)1、求f(x)的最小正周期2、求单调递增区间3、若x∈[0,π/2],f(x)的最小值是-2,求a的值
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![已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a(a∈R,a为常数)1、求f(x)的最小正周期2、求单调递增区间3、若x∈[0,π/2],f(x)的最小值是-2,求a的值](/uploads/image/z/2738999-47-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%282x%2B%CF%80%2F6%29%2Bsin%282x-%CF%80%2F6%29%2Bcos2x%2Ba%28a%E2%88%88R%2Ca%E4%B8%BA%E5%B8%B8%E6%95%B0%EF%BC%891%E3%80%81%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F2%E3%80%81%E6%B1%82%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B43%E3%80%81%E8%8B%A5x%E2%88%88%5B0%2C%CF%80%2F2%5D%2Cf%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E6%98%AF-2%2C%E6%B1%82a%E7%9A%84%E5%80%BC)
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已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a(a∈R,a为常数)1、求f(x)的最小正周期2、求单调递增区间3、若x∈[0,π/2],f(x)的最小值是-2,求a的值
已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a(a∈R,a为常数)
1、求f(x)的最小正周期
2、求单调递增区间
3、若x∈[0,π/2],f(x)的最小值是-2,求a的值
已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a(a∈R,a为常数)1、求f(x)的最小正周期2、求单调递增区间3、若x∈[0,π/2],f(x)的最小值是-2,求a的值
1.
f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a
=√3/2sin2x+1/2cos2x+√3/2sin2x-1/2cos2x+cos2x+a
=√3sin2x+cos2x+a
=2sin(2x+π/6)+a
所以f(x)的最小正周期2π/2=π
2.
sinx的单调递增区间[2kπ-π/2,2kπ+π/2]
所以2kπ-π/2
f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a
=2sin2xcosπ/6+cos2x+a
=2sin(2x+π/6)+a
1)T=π
2)sinx单调递增区间为[2kπ-1/2π,2kπ+1/2π]
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