$\cos x-\sin x=\dfrac{1}{5}$
$\Leftrightarrow \cos^2x+\sin^2x-2\sin x\cos x=\dfrac{1}{25}$
$\Leftrightarrow \sin x\cos x=\dfrac{12}{25}$
Chia 2 vế cho $\sin^2x$:
$\cot x=\dfrac{12}{25}(1+\cot^2x)$ (vì ta có $\dfrac{1}{\sin^2x}=1+\cot^2x$)
$\Leftrightarrow \dfrac{12}{25}\cot^2x-\cot x+\dfrac{12}{25}=0$
$\Leftrightarrow \cot x=\dfrac{4}{3}$ hoặc $\dfrac{3}{4}$