8-4sin^2х=sin2x * ctgx - 9cosx sinх не равен 0 х не равен Пk

Решите задачу:

$8-4\sin^2x=\sin2x\cdot ctgx-9\cos x\\ D(f):\ ctgx: \ x\neq\pi k,\ k\in Z\\ 8-4\sin^2x=2\sin x\cos x\cdot\frac{\cos x}{\sin x}-9\cos x\\ 8-4\sin^2x=2\cos^2x-9\cosx;\\ 8-4+4\cos^2x=2\cos^2x-9\cos x;\\ 4\cos^2x-2\cos^2x+9\cos x+8-4=0;\\ 2\cos^2x+9\cos x+4=0;\\ \cos x=t;\ \ \ -1\leq t\leq1;\\ 2t^2+9t+4=0;\\ D=9^2-4\cdot2\cdot4=81-32=49=(\pm7)^2;\\ t_1=\frac{-9-7}{2\cdot2}=\frac{-16}{4}=-4\notin D(f);\\ t_2=\frac{-9+8}{2\cdot2}=\frac{-2}{4}=-\frac12\tin D(f);\\ \cos x=-\frac12;\\$
$x=\pi-\frac{\pi}{3}+2\pi n, n\in Z$