解题思路:令y=g(x)=2sinx,利用函数y=Asin(ωx+φ)的图象变换可求得f(x)=2sin(2x-[π/6]),从而可求得x∈[0,[π/2]]时,函数y=f(x)的值域.
令y=g(x)=2sinx,
则g(x-[π/6])=2sin(x-[π/6]),
∴f(x)=2sin(2x-[π/6]),
∵x∈[0,[π/2]],
∴2x-[π/6]∈[-[π/6],[5π/6]],
∴sin(2x-[π/6])∈[-[1/2],1],
∴2sin(2x-[π/6])∈[-1,2],
即函数y=f(x)的值域为[-1,2].
故答案为:[-1,2].
点评:
本题考点: 函数y=Asin(ωx+φ)的图象变换.
考点点评: 本题考查函数y=Asin(ωx+φ)的图象变换,考查正弦函数的单调性质,考查运算求解能力,属于中档题.