解由y=cos²x+2sinx
=1-sin^2x+2sinx
=-sin^2x+2sinx+1
令t=sinx
由-π/6≤x≤5π/6
即-1/2≤sinx≤1
即-1/2≤t≤1
故y=-sin^2x+2sinx+1
變?yōu)閥=-t^2+2t+1 t屬于[-1/2,1]
故y=-t^2+2t+1
=-（t-1）^2+2
故當(dāng)t=1時(shí),y有最大值2
當(dāng)t=-1/2時(shí),y有最小值-1/4
故原函數(shù)的值域?yàn)閇-1/4,2].