If $2{\sin }^{2}x-5\sin x\cos x-8{\cos }^{2}x=-2$ then

$\text{ Dividing both sides by }{\cos }^{2}x,$

$\text{ we get }2{\tan }^{2}x-5\tan x-8=-2{\sec }^{2}x$

$\text{ or }2{\tan }^{2}x-5\tan x-8+2\left(1+{\tan }^{2}x\right)=0$

$\text{ Now, }\tan x-2=0\Rightarrow \tan x=2=\tan \alpha$

$\Rightarrow x=n\pi +\alpha =n\pi +{\tan }^{-1}2,n\in Z$

$\text{ or }4\tan x+3=0\text{ or }\tan x=-\frac{3}{4}=\tan \beta \text{ (let) }$

$\Rightarrow x=m\pi +{\tan }^{-1}\left(-\frac{3}{4}\right),\text{ where }m\in Z$