Đáp án:
$\begin{array}{l}
2{\sin ^2}x - 5\sin x.\cos x - {\cos ^2}x = - 2\\
+ Khi:\cos x = 0 \Rightarrow x = \dfrac{\pi }{2} + k\pi \\
\Rightarrow 2{\sin ^2}x = - 2\\
\Rightarrow {\sin ^2}x = - 1\left( {ktm} \right)\\
+ Khi:\cos x \ne 0 \Rightarrow x \ne \dfrac{\pi }{2} + k\pi \\
\Rightarrow \dfrac{{2{{\sin }^2}x}}{{{{\cos }^2}x}} - \dfrac{{5\sin x.\cos x}}{{{{\cos }^2}x}} - 1 = \dfrac{{ - 2}}{{{{\cos }^2}x}}\\
\Rightarrow 2{\tan ^2}x - 5\tan x - 1 = - 2.\left( {{{\tan }^2}x + 1} \right)\\
\Rightarrow 4{\tan ^2}x - 5\tan x + 1 = 0\\
\Rightarrow \left( {4\tan x - 1} \right)\left( {\tan x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \left( {tmdk} \right)\\
x = \arctan \dfrac{1}{4} + k\pi
\end{array} \right.
\end{array}$
