If $5\left({\tan }^{2}x-{\cos }^{2}x\right)=2\cos 2x+9$ then the value of $\cos 4x\text{ is }$

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

Let $u={\tan }^{2}x$we have

$\begin{array}{l}5\left(u-\frac{1}{1+u}\right)=2\left(\frac{1-u}{1+u}\right)+9\\ \Rightarrow 5\left(u^2+u-1\right)=2-2u+9+9:\\ \therefore 5u^2-2u-16=0\Rightarrow (5u+8)(u-2)=0\end{array}$

But $u$ is positive $\therefore u=2$

Now, ${\tan }^{2}x=2$

$\begin{array}{l}\Rightarrow \cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}=\frac{1-2}{1+2}=\frac{-1}{3}\\ \Rightarrow \cos 4x=2{\cos }^{2}2x-1=2\left(\frac{1}{9}\right)-1=\frac{-7}{9}\end{array}$