题目

已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cosx+a,
,1.求最小正周期
2.求f（x）的单调增区间
3 求f（x）的最大值及此时x值
4.若当x∈【0,2/派】时,f（x）的最小值为2,求a

提问时间：2021-03-17

答案

f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a
= sin2xcosπ/6+cos2xsinπ/6 + sin2xcosπ/6-cos2xsinπ/6 + cos2x + a
= 2sin2xcosπ/6 + cos2x + a
= √3sin2x+cos2x+a
= 2(sin2xcosπ/6+cos2xsinπ/6) + a
= 2sin(2x+π/6) + a
最小正周期 = 2π/2 = π
2x+π/6∈（2kπ-π/2,2kπ+π/2）时单调增,∴单调递增区间（kπ-π/3,kπ+π/6）,其中k∈Z
最大值2+a,此时2x+π/6=2kπ+π/2,即x=kπ+π/6,其中k∈Z
x∈[0,π/2]时,2x+π/6∈[-π/6,5π/6],2x+π/6=-π/6时最小值2sin(-π/6)+a=-1+a=2,a=3