\({\cos ^2}\left( {\pi + x} \right) + \sin \left( {\dfrac{\pi }{2} + x} \right)-2 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {-\cos x} \right)^2} + \cos x-2 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\cos ^2}x + \cos x-2 = 0.\)
Пусть \(\cos x = t,\,\,\,\,t\, \in \,\left[ {-1;1} \right]\). Тогда последнее уравнение примет вид:
\({t^2} + t-2 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{t = -2\,\, \notin \,\left[ {-1;1} \right].}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\cos x = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = 2\pi k,\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(2\pi k;\quad k \in Z.\)