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