\({\sin ^2}3x + 5\sin \left( {3x + 3\pi } \right)\cos 3x + 6{\sin ^2}\left( {\dfrac{{3\pi }}{2} + 3x} \right) = 0.\)
Используя формулы приведения, получим:
\(\sin \left( {3x + 3\pi } \right) = -\sin 3x;\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\sin ^2}\left( {\dfrac{{3\pi }}{2} + 3x} \right) = {\left( {-\cos 3x} \right)^2} = {\cos ^2}3x.\)
Тогда получим однородное тригонометрическое уравнение второй степени:
\({\sin ^2}3x-5\sin 3x\cos 3x + 6{\cos ^2}3x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\rm{t}}{{\rm{g}}^2}3x-5{\rm{tg}}3x + 6 = 0.\)
Пусть \({\rm{tg3}}x = t,\,\,\,\,\,t\, \in \,R\). Тогда:
\({t^2}-5t + 6 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 2,\,\,\,\,\,}\\{t = 3.\,\,\,\,}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{{\rm{tg}}3x = 2,}\\{{\rm{tg}}3x = 3\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{3x = {\rm{arctg}}2 + \pi k,}\\{3x = {\rm{arctg}}3 + \pi k\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{1}{3}{\rm{arctg}}2 + \dfrac{{\pi k}}{3},}\\{x = \dfrac{1}{3}{\rm{arctg}}3 + \dfrac{{\pi k}}{3},}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(\dfrac{1}{3}{\rm{arctg}}\,2 + \dfrac{{\pi k}}{3};\;\,\,\;\dfrac{1}{3}{\rm{arctg}}\,3 + \dfrac{{\pi k}}{3};\;\,\,\;k \in Z.\)