\({\rm{2}}\,{\rm{ctg}}\,x-3\,{\rm{tg}}\,x + 5 = 0\)
Так как \({\rm{tg}}\,x \cdot {\rm{ctg}}\,x = 1\), то \({\rm{ctg}}\,x = \dfrac{1}{{{\rm{tg}}\,x}}\). Тогда уравнение примет вид:
\(\dfrac{2}{{{\rm{tg}}\,x}}-3{\rm{tg}}\,x + 5 = 0.\)
Пусть \({\rm{tg}}\,x = t,\,\,\,\,\,t\, \in \,R\). Тогда:
\(\dfrac{2}{t}-3t + 5 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{3{t^2}-5t-2 = 0,}\\{t \ne 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 2,\,\,\,\,\,}\\{t = -\dfrac{1}{3}.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{{\rm{tg}}\,x = 2,\,\,\,}\\{{\rm{tg}}\,x = -\dfrac{1}{3}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = {\rm{arctg}}2 + \pi k,\,\,\,\,\,}\\{x = -{\rm{arctg}}\frac{1}{3} + \pi k,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \({\rm{arctg}}\,{\rm{2}} + \pi k;\quad -{\rm{arctg}}\dfrac{1}{3} + \pi k;\,\,\,\,\,k \in Z.\)