If $5\left({\tan}^{2}x-{\cos}^{2}x\right)=2\cos2x+9$, then the value of $\cos4x$ is:

We know that $\tan ^{2} x=\sec ^{2} x-1$ and $\cos 2 x=2 \cos ^{2} x-1$

$\therefore 5\left(\sec ^{2} x-1-\cos ^{2} x\right)=2\left[2 \cos ^{2} x-1\right]+9$

Take $\cos ^{2} x=t$

$\begin{array}{l} \Rightarrow 5\left(\frac{1}{t}-1-t\right)=2(2 t-1)+9 \\ \Rightarrow \frac{5}{t}-5-5 t=4 t+7 \\ \Rightarrow 5-5 t-5 t^{2}=4 t^{2}+7 t \\ \Rightarrow 9 t^{2}+12 t-5=0 \\ \Rightarrow(3 t+5)(3 t-1)=0 \\ \Rightarrow t=\frac{-5}{3} \text { or } t=\frac{1}{3} \end{array}$

$t \neq \frac{-5}{3}$ as the range of $\cos ^{2} x$ is $[0,1]$

Hence, $\cos ^{2} x=\frac{1}{3}$

$\text { Now, } \begin{aligned} \cos 4 x & =2 \cos ^{2} 2 x-1 \\ & =2\left[2 \cos ^{2} x-1\right]^{2}-1 \\ & =2\left[2\left(\frac{1}{3}\right)-1\right]^{2}-1 \\ & =\frac{2}{9}-1 \\ & =\frac{-7}{9} \end{aligned}$

So, option (D) is correct.