易得：0≦x≦1
不妨令：x=sin²θ,θ∈[0,π/2]
则：3-3x=3(1-x)=3cos²θ
所以,y=sinθ+(√3)cosθ,θ∈[0,π/2]
由辅助角公式：y=2sin(θ+π/3),
θ∈[0,π/2]
则θ+π/3∈[π/3,5π/6]
则：sin(θ+π/3)∈[1/2,1]
所以,y∈[1,2]
即所求值域为[1,2]