已知函数f[x]=sin[x+π
(1)f[x]=sin[x+π/6]+sin[x-π/6]+cosx+a
=sinxcos[π/6]+cosxsin[π/6]+sinxcos[π/6]-cosxsin[π/6]+cosx+a
=2sinxcos[π/6]+cosx+a
=√3sinx+cosx+a
=2sin[x+π/6]+a
当sin[x+π/6]=1,f[x]取到最大值为f[x]max=2+a=1
a=-1
(2)f[x]=2sin[x+π/6]-1
令u=x+π/6，则sinu 的单调增区间是[-π/2+2kπ，π/2+2kπ] (k∈Z)
则sin(x+π/6]的单调增区间是[-π/2+2kπ-π/6,π/2+2...全部
(1)f[x]=sin[x+π/6]+sin[x-π/6]+cosx+a
=sinxcos[π/6]+cosxsin[π/6]+sinxcos[π/6]-cosxsin[π/6]+cosx+a
=2sinxcos[π/6]+cosx+a
=√3sinx+cosx+a
=2sin[x+π/6]+a
当sin[x+π/6]=1,f[x]取到最大值为f[x]max=2+a=1
a=-1
(2)f[x]=2sin[x+π/6]-1
令u=x+π/6，则sinu 的单调增区间是[-π/2+2kπ，π/2+2kπ] (k∈Z)
则sin(x+π/6]的单调增区间是[-π/2+2kπ-π/6,π/2+2kπ-π/6](k∈Z)
则f[x]的单调增区间为[-2π/3+2kπ，π/3+2kπ] ，(π∈Z)
（3）f[x]=2sin[x+π/6]-1>=0
sin[x+π/6]>=1/2
如图在[0,2π]一个周期里满足sin[x+π/6]>=1/2的x范围是
π/6=0成立的X的取值集为{x|2kπ<=x<=2π/3+2kπ，(k∈Z)}。
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