已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a(a∈R,a为常数)
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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
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)](/uploads/image/z/1043550-54-0.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%89)
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
=√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
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