\(\sqrt {3{\rm{ctg}}\,x-2} \le {\rm{ctg}}\,x.\)
Пусть \({\rm{ctg}}\,x = t.\) Тогда:
\(\sqrt {3t-2} \le t\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{3t-2 \ge 0,}\\{t \ge 0,\,\,\,\,\,\,\,\,\,}\\{3t-2 \le {t^2}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{t \ge \dfrac{2}{3},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{t \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{t^2}-3t + 2 \ge 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{t \ge \dfrac{2}{3},}\\{\left[ {\begin{array}{*{20}{c}}{t \le 1,}\\{t \ge 2}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,t\, \in \,\left[ {\dfrac{2}{3};1} \right] \cup \left[ {2;\infty } \right).\)
Возвращаясь к прежней переменной, получим:
\(\left[ {\begin{array}{*{20}{c}}{\dfrac{2}{3} \le {\rm{ctg}}\,x \le 1,}\\{{\rm{ctg}}\,x \ge 2\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x\, \in \,\left[ {\dfrac{\pi }{4} + \pi k;{\rm{arcctg}}\dfrac{2}{3} + \pi k} \right],\,\,}\\{x\, \in \,\left( {\pi k;{\rm{arcctg}}\,2 + \pi k} \right],\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(\left( {\pi k;{\rm{arcctg}}\,2 + \pi k} \right] \cup \left[ {\dfrac{\pi }{4} + \pi k;{\rm{arcctg}}\dfrac{2}{3} + \pi k} \right],\,\,\,k\, \in \,Z.\)