若tan(2x-π/3)<=1,则x的取值范围是？

详细过程 谢谢各位呀 答案应该是kπ/2-π/12<x<=kπ/2+7π/24

tan(2x-π/3)<=1
(kπ-π/2)<(2x-π/3)<=(kπ+π/4) 自己画一个周期的图像就明白了
kπ/2-π/12<x<=kπ/2+7π/24

首先要知道tan函数的周期为π，于是我们可以先考虑一个周期(-π/2,π/2)（端点处tan函数无意义）,然后加上两边kπ就可以了。若tan(2x-π/3)≤1则应有-π/2＜2x-π/3≤π/4,得到-π/12＜x≤7π/24,然后延拓到整个x轴上就有kπ-π/12＜x≤kπ+7π/24。 最后要声明的是你原来的答案是不对的，原来两边不应该加kπ/2，而应该是kπ。

tan(2x-π/3)<=1 所以-π/2+kπ<2x-π/3<=π/4+kπ -π/6+kπ<2x<=7π/12+kπ -π/24+kπ<x<=7π/24+kπ

tan(2x-π/6)≤1 2kπ-π/22kπ+π/2 2kπ-π/32kπ+2π/3 kπ-π/6kπ+π/3