f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a=2sinx·cosπ/6+cosx+a=根号3sinx+cosx+a=2sin(x+π)+a (1)由于f(x)max=2+a=1,所以a=-1 (2)因为f(x)≥0 所以 2 sin( x + π /6)-1≥ 0 即sin(x+π/6)≥ 1/2,所以π/6+2kπ≤x+π/6≤5π/6+2kπ 　　f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a=2sinx·cosπ/6+cosx+a=根号3sinx+cosx+a=2sin(x+π)+a (1)由于f(x)max=2+a=1,所以a=-1 (2)因为f(x)≥0 所以 2 sin( x + π /6)-1≥ 0 即sin(x+π/6)≥ 1/2,所以π/6+2kπ≤x+π/6≤5π/6+2kπ