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=-1f(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...