\(2{\sin ^2}2x-3\sin 2x\cos 2x + {\cos ^2}2x = 0.\)
Однородное тригонометрическое уравнение второй степени:
\(2{\sin ^2}2x-3\sin 2x\cos 2x + {\cos ^2}2x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2{\rm{t}}{{\rm{g}}^2}2x-3{\rm{tg}}2x + 1 = 0\)
Пусть \({\rm{tg}}2x = t,\,\,\,\,\,t\, \in \,R\). Тогда уравнение примет вид:
\(2{t^2}-3t + 1 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,\,\,}\\{t = \dfrac{1}{2}.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{{\rm{tg}}2x = 1,}\\{{\rm{tg}}2x = \dfrac{1}{2}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x = \dfrac{\pi }{4} + \pi k,\,\,\,\,\,\,\,\,\,\,\,}\\{2x = {\rm{arctg}}\dfrac{1}{2} + \pi k}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{8} + \dfrac{{\pi k}}{2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = \dfrac{1}{2}{\rm{arctg}}\dfrac{1}{2} + \dfrac{{\pi k}}{2},}\end{array}} \right.\,\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(\dfrac{\pi }{8} + \dfrac{{\pi k}}{2};\,\,\,\,\,\,\dfrac{1}{2}{\rm{arctg}}\,\dfrac{1}{2} + \dfrac{{\pi k}}{2};\,\,\,\,\,k \in Z.\)