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

f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a=√3sinx+cosx+a=2sin(x+π/6)+a
因为f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a的最大值为1,所以,a=-1
f(x)=√3sinx+cosx+a=2sin(x+π/6)-1大于等于0,所以2sin(x+π/6)大于等于1
所以x的范围是[2kπ,2kπ+2π/3]

f(x)=sin(x+TT/6)+sin(x-TT/6)+cosx+a
=2sinxcos(TT/6)+cosx+a
=√3sinx+cosx+a
=2sin(x+TT/6)+a最大值为1
2+a=1 =>a=-1
f(x)=2sin(x+TT/6)-1>=0
sin(x+TT/6)>=1/2
(2k+1/6)TT<=x+TT/6=<(2k+5/6)TT
2kTT<=x=<(2k+2/3)TT