Question

一道高一数学题（请进！请详细说明！谢谢！）

若tan(2x-π/6)≤1,则x的取值范围__________
（我大概知道怎么做，但是做不对，大家帮帮忙!）
哦……答案是（kπ/2-π/6<x≤kπ/2+5π/24,k∈Z）

Answer 1

-π/2+Kπ<（2x-π/6)<=π/4+Kπ
解：
-1/3π+Kπ<2x<=5/12π+Kπ
1/6π+K/2π<x<=5/24π+K/2π

Answer 2

tan(2x-π/6)≤1
2kπ-π/2<2x-π/6≤2kπ+π/4
2kπ+π/2<2x-π/6≤2kπ+5π/4

2kπ-π/3<2x≤2kπ+5π/12
2kπ+2π/3<2x≤2kπ+17π/12

kπ-π/6<x≤kπ+5π/24
kπ+π/3<x≤kπ+17π/24

Answer 3

解：tan(2x-π/6)≤1
2kπ-π/2<2x-π/6≤2kπ+π/4
2kπ+π/2<2x-π/6≤2kπ+5π/4

2kπ-π/3<2x≤2kπ+5π/12
2kπ+2π/3<2x≤2kπ+17π/12

kπ-π/6<x≤kπ+5π/24
kπ+π/3<x≤kπ+17π/24

所以kπ/2-π/6<x≤kπ/2+5π/24 k∈Z