\(\dfrac{{{\rm{tg}}\,2x + {\rm{tg}}\,x}}{{1-{\rm{tg}}\,2x\,{\rm{tg}}\,x}} = \sqrt 3 \)
Воспользуемся формулой тангенса суммы:
\({\rm{tg}}\left( {\alpha + \beta } \right) = \dfrac{{{\rm{tg}}\alpha + {\rm{tg}}\beta }}{{1-{\rm{tg}}\alpha {\rm{tg}}\beta }}.\)
\(\dfrac{{{\rm{tg}}2x + {\rm{tg}}x\,}}{{1-{\rm{tg}}2x\,\,{\rm{tg}}x\,}} = \sqrt 3 \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\rm{tg}}\left( {2x + x} \right) = \sqrt 3 \,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\rm{tg}}3x = \sqrt 3 \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,3x = \dfrac{\pi }{3} + \pi k\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \dfrac{\pi }{9} + \dfrac{{\pi k}}{3},\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(\dfrac{\pi }{9} + \dfrac{{\pi k}}{3};\quad \,k \in Z.\)