换元法.
令sinx=t,
∵ x∈[0,π/2]
∴ t∈[0,1]
∴ f(x)=2cos²x+sinx-1
=2-2sin²x+sinx-1
=-2sin²x+sinx+1
=-2t²+t+1
=-2(t-1/4)²+9/8
∴ t=1/4时,f(x)有最大值9/8
t=1时,f(x)有最小值0
∴ f(x)的值域是[0,9/8]