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