已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a的最大值为1（1）求常数a的值.（2）求使f(

解1：
f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a
f(x)=2sinxcos(π/6)+cosx+a
f(x)=√3sinx+cosx+a
f(x)=2[(√3/2)sinx+(1/2)cosx]+a
f(x)=2[cos(π/6)sinx+sin(π/6)cosx]+a
f(x)=2sin(x+π/6)+a
可见：f(x)≤2+a
因为：f(x)≤1
所以：2+a=1
解得：a=-1
解2：
由上解,有：f(x)=2sin(x+π/6)-1
因为：f(x)≥0
所以：2sin(x+π/6)-1≥0
即：sin(x+π/6)≥1/2
有：2kπ+5π/6≥x+π/6≥2kπ+π/6,k∈Z
解得：2kπ+2π/3≥x≥2kπ,k∈Z

f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a
=2sin(x+π/4)+a
最大值2+a=1 a=-1
2sin(x+π/4)-1>=0
sin(x+π/4)>=1/2
2kπ-π/12<=x<=2kπ+7π/12

y=2sinxcosπ/6+cosx+a=根号3sinx+cosx+a=2sin(x+π/6)+a
2sin(x+π/6)的最大值为2因为ymax=2+a=1
所以a=-1
所以y>=0->sin(x+π/6)>=1/2
所以2kπ+π/6<=x+π/6<=2kπ+5π/6
所以2kπ<=x<=2kπ+2π/3
集合为{x|2kπ<=x<=2kπ+2π/3,k属于Z}

(1) f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a
=2sinxcosπ/6+cosx+a
=√3sinx+cosx+a
=2sin(x+π/4)+a
最大值=2+a=1 a=-1
(2)
2sin(x+π/4)-1>=0
sin(x+π/4)>=1/2
2kπ+π/6<=x+π/4<=2kπ+5π/6
2kπ-π/12<=x<=2kπ+7π/12