3+2cosx+sinx/1+2cosx+3sinx

y=(3+2cos+sinx)/(1+2cosx+3sinx)
→(3y-1)sinx+(2y-2)cosx=3-y.
构造向量m=(3y-1,2y-2),n=(sinx,cosx).
则 |m*n|≤|m|*|n|
→[(3y-1)^2+(2y-2)^2]*[(sin)^2+(cosx)^2]≤[(3y-1)sinx+(2y-2)cosx]^2
→[(3y-1)^2+(2y-2)^2]*1≥(3-y)^2
→12y^2-8y-4≥0
解得,y≤-1/3,或y≥1.
即所求值域为 (-∞,-1/3]∪[1,+∞).