\(\dfrac{{{\rm{tg}}\,x + 5}}{3} = \dfrac{1}{{{{\cos }^2}x}}\)
Воспользуемся формулой: \(1 + {\rm{t}}{{\rm{g}}^2}x = \dfrac{1}{{{{\cos }^2}x}}\). Тогда уравнение примет вид:
\(\dfrac{{{\rm{tg}}\,x + 5}}{3} = 1 + {\rm{t}}{{\rm{g}}^2}x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,3{\rm{t}}{{\rm{g}}^2}x-{\rm{tg}}\,x-2 = 0.\)
Пусть \({\rm{tg}}\,x = t,\,\,\,\,\,t\, \in \,R\). Тогда:
\(3{t^2}-t-2 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,\,\,\,\,\,\,}\\{t = -\dfrac{2}{3}.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{{\rm{tg}}\,x = 1,\,\,\,\,}\\{{\rm{tg}}\,x = -\dfrac{2}{3}}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{4} + \pi k,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = -{\rm{arctg}}\dfrac{2}{3} + \pi k,}\end{array}} \right.} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(\dfrac{\pi }{4} + \pi k;\quad -{\rm{arctg}}\dfrac{2}{3} + \pi k;\,\,\,\,\,k \in Z.\)