\hfill\break 【分析】\\ 由题意可得$t=\omega x+\frac{\text{ }\pi\text{ }}{\text{6}}\in (\frac{\text{ }\pi\text{ }\omega }{2}+\frac{\text{ }\pi\text{ }}{\text{6}},\text{ }\pi\text{ }\omega +\frac{\text{ }\pi\text{ }}{\text{6}})$，再根据$y=\sin t$的单调区间，列出不等式组求解即可.\\ 【详解】\\ 解：因为$\omega \textgreater0$，$x\in \left( \frac{\text{ }\pi\text{ }}{\text{2}}\text{, }\pi\text{ } \right)$，\\ 所以$t=\omega x+\frac{\text{ }\pi\text{ }}{\text{6}}\in (\frac{\text{ }\pi\text{ }\omega }{2}+\frac{\text{ }\pi\text{ }}{\text{6}},\text{ }\pi\text{ }\omega +\frac{\text{ }\pi\text{ }}{\text{6}})$，\\ 又因为$y=\sin t$在$(\frac{\text{ }\pi\text{ }}{2}+2k\text{ }\pi\text{ },\frac{3\text{ }\pi\text{ }}{2}+2k\text{ }\pi\text{ }),k\in \text{Z}$上单调递减，\\ 所以$\text{ } {\text{ }\begin{array}{*{35}{l}} \frac{\text{ }\pi\text{ }\omega }{2}+\frac{\text{ }\pi\text{ }}{\text{6}}\ge \frac{\text{ }\pi\text{ }}{2}+2k\text{ }\pi\text{ }    \text{ }\pi\text{ }\omega +\frac{\text{ }\pi\text{ }}{\text{6}}\le \frac{3\text{ }\pi\text{ }}{2}+2k\text{ }\pi\text{ }    \end{array}\text{ }$，$k\in \text{Z}$,\\ 解得：$\frac{2}{3}+2k\text{ }\pi\text{ }\le \omega \le \frac{4}{3}+2k\text{ }\pi\text{ },k\in \text{Z}$.\\ 当$k=0$时，$\frac{2}{3}\le \omega \le \frac{4}{3}$.\\ 故选：A 
已知$\omega \textgreater0$，函数$f\left( x \right)=\text{sin}\left( \omega x+\frac{\text{ }\pi\text{ }}{\text{6}} \right)$在$\left( \frac{\text{ }\pi\text{ }}{\text{2}},\text{ }\pi\text{ } \right)$上单调递减，则$\omega $的取值范围是（~~~~~~~）

1: $\left[ \frac{2}{3},\frac{4}{3} \right]$ 
2: $\left[ \frac{2}{3},\frac{3}{4} \right]$ 
3: $\left( 0,\frac{2}{3} \right]$ 
4: $\left( 0,\frac{3}{2} \right]$ 