求函数y=2sin^2(x)+5cosx-1的值域

y=2(1-cos^2x)+5cosx-1
=-2cos^2x+5cosx+1
=-2(cos^2x-5/2cosx-1/2)
=-2(cos^2x-5/2cosx+25/16-33/16)
=-2(cosx-5/4)^2+33/8
∵cosx∈[-1，1]
所以cosx-5/4∈[-9/4,-1/4]
y∈[-6,4]

y=2sin^2(x)+5cosx-1
=2-2cos^2(x)+5cosx-1
=-2cos^2(x)+5cosx+1
=-2(cosx-5/4)^2+25/8+1
=-2(cosx-5/4)^2+33/8

cosx-5/4≤1-5/4=1/4，则(cosx-5/4)^2≥（1/4）^2=1/16
则y=-2(cosx-5/4)^2+33/8≤-1/8+33/8=4；
cosx-5/4≥-1-5/4=-9/4，则(cosx-5/4)^2≤（-9/4）^2=81/16
则y=-2(cosx-5/4)^2+33/8≥-81/8+33/8=-6.

即值域为[-6,4]