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