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