How find the equation $\cot x=\frac{\sin 20^\circ - \sin 80^\circ \cos 20^\circ}{\sin 80^\circ \sin 20^\circ}$ let $x\in R$, and such 
$$\cot x =\frac{\sin 20^\circ -\sin 80^\circ \cos 20^\circ}{\sin 80^\circ \sin 20^\circ}$$
Find $x$
my idea:
$$\cot x=\csc 80^\circ - \cot 20^\circ$$
then I can't 

Let $\theta=20^\circ$, and defin
$\Delta=\sin\theta-\sin4\theta\cos\theta+\sqrt{3}\sin4\theta\sin\theta
$.
 Then
$$\eqalign{2\Delta&=2\sin\theta-(\sin5\theta+\sin3\theta)+\sqrt{3}(\cos3\theta-\cos5\theta)\cr
&=2\sin\theta-\sin5\theta-\sqrt{3}\cos5\theta\cr
&=2\sin\theta-2(\sin5\theta\cos60^\circ+\cos5\theta\sin60^\circ)\cr
&=2\sin\theta-2\sin8\theta=0
}
$$
From $\Delta=0$ we conclude that $cot x=-\sqrt{3}$ thus $x\in\left\{-\frac{\pi}{6}+\pi k:k\in\Bbb{Z}\right\}$.$\qquad\square$
