If 5 tan 2 ⁡ x - cos 2 ⁡ x = 2 cos ⁡   2 x + 9 , then the value of cos ⁡ 4 x is

$5\left(\frac{1-\cos 2x}{1+\cos 2x}-\frac{1+\cos 2x}{2}\right)=2\cos 2x+9$

Let, $\cos 2x=y$

$\therefore 5(\frac{1-y}{1+y}-\frac{1+y}{2})=2y+9$

$\Rightarrow 5\left[-y^2-4y+1\right]=4y^2+22y+18$

$\Rightarrow 9y^2+42y+13=0$

$\Rightarrow y=-\frac{1}{3}$ or $y=-\frac{13}{3}$ (Not possible)

Now, $\cos 4x=2{\cos }^{2}2x-1$

$\Rightarrow \cos 4x=2{\left(-\frac{1}{3}\right)}^2-1$

$\Rightarrow \cos 4x=-\frac{7}{9}$