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