y=(3cosx+1)/(cosx+2)x∈[-π/2,2π/3]的值域为

问题描述:

y=(3cosx+1)/(cosx+2)x∈[-π/2,2π/3]的值域为

y=(3cosx+1)/(cosx+2)
ycosx+2y=3cosx+1
(y-3)cosx=1-2y
cosx=(1-2y)/(y-3)
x∈[-π/2,2π/3]
-1/2≤cosx≤1
-1/2≤(1-2y)/(y-3)≤1
(a)-1/2≤(1-2y)/(y-3)
(2-4y+y-3)/(2y-6)≥0 (3y+1)/(y-3)≤0
-1/3≤y3或y≤4/3 (2)
综上(1)(2) -1/3≤y≤4/3
即值域为[-1/3,4/3]