срочно подробное решение)

$8cos^4x = 11cos2x - 1\\ 8cos^4x = 11(2cos^2x -1) - 1\\ 8cos^4x = 22cos^2x - 11 -1\\ 8cos^4x - 22cos^2x + 12 = 0\\ cos^2x = t\\ 8t^2 - 22t + 12 = 0\\ D = 484 - 4*8*12 = 100\\ t_{1,2} = \frac{22 \pm 10}{-16} = -2; -\frac{1}{8}.\\ cos^2x = -2\\ -cosx \neq \sqrt{2} \\ cos^2x = -\frac{1}{8}\\ -cos^2x = \frac{1}{8}\\ -cosx = \frac{1}{2\sqrt{2}}\\ cosx = -\frac{1}{2\sqrt{2}}\\ x =\pm arccos(-\frac{1}{2\sqrt{2}}) +2\pi*k, k\in Z$

$tg^2x - 2tgx + 6ctgx = 3\\ tg^2x - 2tgx = 3 - 6ctgx\\ tg^2x - 2tgx = 3(1- 2ctgx)\\ tg^2x - 2tgx = 3(1- 2tg^{-1}x)\\ tg^2x = 3\\ tgx = \sqrt{3}\\ x = \frac{\pi}{3}+\pi*k, k\in Z\\$