\({\sin ^2}x + 3\sin x\cos x-4{\cos ^2}x = 0.\)
Однородное тригонометрическое уравнение второй степени:
\({\sin ^2}x + 3\sin x\cos x-4{\cos ^2}x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\rm{t}}{{\rm{g}}^2}x + 3{\rm{tg}}\,x-4 = 0.\)
Пусть \({\rm{tg}}\,x = t,\,\,\,\,\,t\, \in \,R\). 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: \(\dfrac{\pi }{4} + \pi k;\,\,\,\,\,\,-{\rm{arctg}}\,\,4 + \pi k;\,\,\,\,\,k \in Z.\)