y=cosx+sin（π/6-x),x属于【0,π】的最小值和最大值

y=cosx+sin(π/6)cosx-cos(π/6)sinx=(3/2)cosx-(√3/2)sinx=-√3*[sinx*(1/2)-cosx*(√3/2)]=-√3 [sinx*cos(π/3)-cosx*sin(π/3)]=-√3sin(x-π/3)∵ x∈[0,π]∴ x-π/3∈[-π/3,2π/3]∴ x-π/3=-π/3,即x=0时,...