Объяснение:

[tex]tg\alpha =-\sqrt{5} \ \ \ \ \frac{11\pi }{2} < \alpha < 6\pi \ \ \ \ cos2\alpha =?\\\frac{sin\alpha }{cos\alpha } =-\sqrt{5}\\ sin\alpha =-\sqrt{5} *cos\alpha\\( sin\alpha)^2 =(-\sqrt{5} *cos\alpha)^2\\sin^2\alpha =5*cos^2\alpha \\sin^2\alpha +cos^2\alpha =5*cos^2\alpha +cos^2\alpha\\ 6*cos^2\alpha =1\ |:6\\cos^2\alpha =\frac{1}{6}.\\[/tex]

[tex]sin^2\alpha +cos^2\alpha =1\\sin^2\alpha =1-cos^2\alpha =1-\frac{1}{6} =\frac{5}{6} \\cos2\alpha =cos^2\alpha -sin^2\alpha =\frac{1}{6}-\frac{5}{6} =-\frac{4}{6}=-\frac{2}{3}.[/tex]

Ответ: cos2α=-2/3.