Воспользуемся основным тригонометрическим тождеством:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {\dfrac{{2\sqrt 2 }}{3}} \right)^2} + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\cos ^2}\alpha = \dfrac{1}{9}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\cos \alpha = \dfrac{1}{3},\,\,\,\,}\\{\cos \alpha = -\dfrac{1}{3}.}\end{array}} \right.\)
Так как \(\alpha \, \in \,\left( {2\pi ;\dfrac{{5\pi }}{2}} \right)\) (I четверть), то \(\cos \alpha > 0\), то есть \(\cos \alpha = \dfrac{1}{3}.\)
Воспользуемся формулой косинуса двойного угла:
\(\cos 2\alpha = 2{\cos ^2}\alpha -1.\)
\(\cos \alpha = \dfrac{1}{3}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2{\cos ^2}\dfrac{\alpha }{2}-1 = \dfrac{1}{3}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\cos ^2}\dfrac{\alpha }{2} = \dfrac{2}{3}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\cos \dfrac{\alpha }{2} = \sqrt {\dfrac{2}{3}} ,\,\,\,\,}\\{\cos \dfrac{\alpha }{2} = -\sqrt {\dfrac{2}{3}} .}\end{array}} \right.\)
Так как \(\alpha \, \in \,\left( {2\pi ;\dfrac{{5\pi }}{2}} \right)\), то \(\dfrac{\alpha }{2}\, \in \,\left( {\pi ;\dfrac{{5\pi }}{4}} \right)\) (III четверть), значит \(\cos \dfrac{\alpha }{2} < 0\), то есть \(\cos \dfrac{\alpha }{2} = -\sqrt {\dfrac{2}{3}} .\)
Тогда: \(\sqrt 6 \cos \dfrac{\alpha }{2} = \sqrt 6 \cdot \left( {-\sqrt {\dfrac{2}{3}} } \right) = -2.\)
Ответ: \(-2.\)