若 x∈[π/6,π/3]时,k+tan(2x-π/3)的值总不大于零,求实数k的取值范围.

若 x∈[π/6,π/3]时,k+tan(2x-π/3)的值总不大于零,求实数k的取值范围.

当x∈[π/6,π/3]时,2x-π/3∈[0,π/3]
所以tan(2x-π/3)∈[0,√3〕
因为k+tan(2x-π/3)≤0
所以k+√3≤0
所以k≤－√3