\(\dfrac{{{\rm{ctg}}\,6x\,\,{\rm{ctg}}\,4x + 1}}{{{\rm{ctg}}\,4x\,-{\rm{ctg}}\,6x}} = 1\)
Воспользуемся формулой котангенса разности:
\({\rm{ctg}}\left( {\alpha -\beta } \right) = \dfrac{{{\rm{ctg}}\alpha {\rm{ctg}}\beta + 1}}{{{\rm{ctg}}\beta -{\rm{ctg}}\alpha }}.\)
\(\dfrac{{{\rm{ctg}}6x\,{\rm{ctg}}4x + 1}}{{{\rm{ctg}}4x-{\rm{ctg}}6x}} = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{\rm{ctg}}\left( {6x-4x} \right) = 1\,\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\rm{ctg}}2x = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2x = \dfrac{\pi }{4} + \pi k\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \dfrac{\pi }{8} + \dfrac{{\pi k}}{2},\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(\dfrac{\pi }{8} + \dfrac{{\pi k}}{2};\quad \,k \in Z.\)