\(\left( {4{{\cos }^2}x-3} \right)\left( {{\rm{ctg}}\,x-1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{4{{\cos }^2}x-3 = 0,}\\{{\rm{ctg}}\,x-1 = 0\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\cos }^2}x = \dfrac{3}{4},}\\{{\rm{ctg}}\,x = 1\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow } \right.} \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\cos x = \dfrac{{\sqrt 3 }}{2},\,\,\,\,}\\{\cos x = -\dfrac{{\sqrt 3 }}{2},}\\{{\rm{ctg}}\,x = 1\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \pm \dfrac{\pi }{6} + 2\pi k,\,\,\,}\\{x = \pm \dfrac{{5\pi }}{6} + 2\pi k,}\\{x = \dfrac{\pi }{4} + \pi k\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \pm \dfrac{\pi }{6} + \pi k,}\\{x = \dfrac{\pi }{4} + \pi k,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \( \pm \dfrac{\pi }{6} + \pi k;\quad \dfrac{\pi }{4} + \pi k;\quad \,k \in Z.\)