$2sin^2x-5sinxcosx-cos^2x=-2(1)$
$x=\dfrac{\pi}{2}+k\pi(k∈Z)⇔\begin{cases}cosx=0\\sin^2x=1\end{cases}$
$⇔2=-2$ (vô lý)
$⇒x=\dfrac{\pi}{2}+k\pi(k∈Z)$ không phải là nghiệm
$x\ne\dfrac{\pi}{2}+k\pi(k∈Z)$
$⇔2tan^2x-5tanx-1=-2(1+tan^2x)$
$⇔4tan^2x-5tanx+1=0$
$⇔tanx=1$
$⇔x=\dfrac{\pi}{4}+k\pi(k∈Z)$
$tanx=\dfrac{1}{4}$
$⇔x=arctan(\dfrac{1}{4})+k\pi(k∈Z)$