f(x)=cosx+sin^2x
=cosx+1-cos^2x
=-cos^2x+cosx+1
f（π/6） =-cos^2（π/6）+cos（π/6）+1=-3/4+根号3/2+1=根号3/2+1/4
设t=cosx ,t∈[-1,1]
g(t)=-t^2+t+1
=-(t-1/2)^2+1/4+1
=-(t-1/2)^2+5/4
t=1/2时,取得最大值5/4
t=-1时,取得最小值-1
所以f（x）最大值5/4；最小值-1