已知f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a的最大值为1

1、
f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a
=√3sinx+cosx+a
=√[(√3)²+1²]sin(x+z)+a
=2sin(x+z)+a
其中tanz=1/√3,所以z=π/6
sin(x+z)最大=1
所以f(x)最大=2+a=1
a=-1

2、
f(x)=2sin(x+π/6)-1>=0
sin(x+π/6)>=1/2
所以2kπ+π/6<=x+π/6<=2kπ+5π/6
2kπ<=x<=2kπ+2π/3
所以集合是{x|2kπ<=x<=2kπ+2π/3,k是整数}