\({\rm{ct}}{{\rm{g}}^2}x-3{\rm{ctg}}\,x + 2 = 0.\)
Пусть \({\rm{ctg}}\,x = t,\,\,\,\,\,\,t\, \in \,R\). Тогда уравнение примет вид:
\({t^2}-3t + 2 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,\,}\\{t = 2.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{{\rm{ctg}}\,x = 1,}\\{{\rm{ctg}}\,x = 2}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{4} + \pi k,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = {\rm{arcctg}}2 + \pi k,}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.} \right.\)
Ответ: \(\,\dfrac{\pi }{4} + \pi k;\quad \,{\rm{arcctg}}\,{\rm{2}} + \pi k;\quad k \in Z.\)