Пусть \(\cos \alpha = t\), тогда:
\(15{t^2}-2t-1 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = \dfrac{1}{3},\,\,\,}\\{t = -\dfrac{1}{5}.}\end{array}} \right.\)
Следовательно: \(\left[ {\begin{array}{*{20}{c}}{\cos \alpha = \dfrac{1}{3},\,\,\,}\\{\cos \alpha = -\dfrac{1}{5}.}\end{array}} \right.\)
Так как \(-\pi < \alpha < -\dfrac{\pi }{2}\) (III четверть), то \(\cos \alpha < 0\), то есть \(\cos \alpha = -\dfrac{1}{5}.\)
Воспользуемся формулой: \(1 + {\rm{t}}{{\rm{g}}^2}\alpha = \dfrac{1}{{{{\cos }^2}\alpha }}.\)
\(1 + {\rm{t}}{{\rm{g}}^2}\alpha = \dfrac{1}{{{{\left( {-\dfrac{1}{5}} \right)}^2}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\rm{t}}{{\rm{g}}^2}\alpha = 24\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{\rm{tg}}\alpha = \sqrt {24} ,\,\,\,}\\{{\rm{tg}}\alpha = -\sqrt {24} .}\end{array}} \right.\)
Так как \(-\pi < \alpha < -\dfrac{\pi }{2}\) (III четверть), то \({\rm{tg}}\alpha > 0\), то есть \({\rm{tg}}\alpha = \sqrt {24} = 2\sqrt 6 .\)
Тогда: \(\sqrt 6 {\rm{tg}}\alpha = \sqrt 6 \cdot 2 \cdot \sqrt 6 = 12.\)
Ответ: 12.