f(x)=cos(x+π/6)-cos(x-π/6)+SQR（3）*cosx=cosxcosπ/6-sinxsinπ/6-cosxcosπ/6-sinxsinπ/6+SQR（3）*cosx=-sinx+SQR（3）cosx=-2sin(x-π/3）因为x∈[-π/2,0],所以(x-π/3）∈[-5π/6,-π/3],所以函数最小值...