求函数f(x)=sin(2x-π/6)+1/2在[0,2π/3]上的取值范围.
∵ x∈[0,2π/3]
∴ 2x-π/6∈[-π/6,2×2π/3-π/6]=[-π/6,7π/6]
∴ f(x)∈[-1/2,1]