Ta có: $\cot\alpha = 2$
$\Rightarrow \tan\alpha = \dfrac{1}{2}$
$\Rightarrow \cot\alpha + \tan\alpha = 2 + \dfrac{1}{2} = \dfrac{5}{2}$
$\Leftrightarrow \dfrac{\cos^2\alpha + \sin^2\alpha}{\sin\alpha.\cos\alpha} = \dfrac{5}{2}$
$\Leftrightarrow \sin\alpha.\cos\alpha = \dfrac{2}{5}$
Ta lại có:
$\sin^2\alpha + \cos^2\alpha = 1$
$\Leftrightarrow (\sin\alpha + \cos\alpha)^2 - 2\sin\alpha.\cos\alpha = 1$
$\Leftrightarrow |\sin\alpha + \cos\alpha| = \sqrt{1 + 2.\dfrac{2}{5}} = \dfrac{3\sqrt5}{5}$
$+) \, \begin{cases}\sin\alpha.\cos\alpha = \dfrac{2}{5}\\\sin\alpha + \cos\alpha= \dfrac{3\sqrt5}{5}\end{cases}$
$\Leftrightarrow \begin{cases}\left[\begin{array}{l}\sin x = \dfrac{\sqrt5}{5}\\\cos x =\dfrac{2\sqrt5}{5}\end{array}\right.\\\left[\begin{array}{l}\sin x = \dfrac{2\sqrt5}{5}\\\cos x =\dfrac{\sqrt5}{5}\end{array}\right.\end{cases}$
$+) \, \begin{cases}\sin\alpha.\cos\alpha = \dfrac{2}{5}\\\sin\alpha + \cos\alpha= -\dfrac{3\sqrt5}{5}\end{cases}$
$\Leftrightarrow \begin{cases}\left[\begin{array}{l}\sin x = -\dfrac{\sqrt5}{5}\\\cos x =-\dfrac{2\sqrt5}{5}\end{array}\right.\\\left[\begin{array}{l}\sin x = -\dfrac{2\sqrt5}{5}\\\cos x =-\dfrac{\sqrt5}{5}\end{array}\right.\end{cases}$
