\(3\cos x + \dfrac{1}{{\cos x-1}} = -\dfrac{1}{2}.\)
Пусть \(\cos x = t,\,\,\,\,t\, \in \,\left[ {-1;1} \right]\). Тогда уравнение примет вид:
\(3t + \dfrac{1}{{t-1}} + \dfrac{1}{2} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{6{t^2}-5t + 1 = 0,}\\{t \ne 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = \dfrac{1}{2},}\\{t = \dfrac{1}{3}.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{\cos x = \dfrac{1}{2},}\\{\cos x = \dfrac{1}{3}\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \pm \dfrac{\pi }{3} + 2\pi k,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = \pm \arccos \dfrac{1}{3} + 2\pi k,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \( \pm \,\dfrac{\pi }{3} + 2\pi k;\quad \, \pm \arccos \dfrac{1}{3} + 2\pi k;\quad k \in Z.\)