\(2{\cos ^2}\dfrac{x}{2} + \cos \dfrac{x}{2}-1 = 0.\)
Пусть \(\cos \dfrac{x}{2} = t,\,\,\,\,\,\,t\, \in \,\left[ {-1;1} \right]\). Тогда уравнение примет вид:
\(2{t^2} + t-1 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = -1,}\\{t = \dfrac{1}{2}.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{\cos \dfrac{x}{2} = -1,}\\{\cos \dfrac{x}{2} = \dfrac{1}{2}\,}\end{array}\,\,\,} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\dfrac{x}{2} = \pi + 2\pi k,\,\,\,\,\,\,}\\{\dfrac{x}{2} = \pm \dfrac{\pi }{3} + 2\pi k\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 2\pi + 4\pi k,\,\,\,\,\,\,}\\{x = \pm \dfrac{{2\pi }}{3} + 4\pi k,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(2\pi + 4\pi k;\quad \pm \dfrac{{2\pi }}{3} + 4\pi k;\quad k \in Z.\)