3+2cosx+sinx/1+2cosx+3sinx
1个回答

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,+∞).