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 
 A: 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 $x\in\left\{-\frac{\pi}{6}+\pi k:k\in\Bbb{Z}\right\}$.$\qquad\square$
A: $$\sin80^\circ=2\sin40^\circ\cos40^\circ=2(2\sin20^\circ\cos20^\circ)\cos40^\circ$$
$$\implies\csc80^\circ-\cot20^\circ=\frac{1-4\cos^220^\circ\cos40^\circ}{\sin80^\circ}$$
Now $\displaystyle N= 1-4\cos^220^\circ\cos40^\circ=1-2\cos40^\circ(2\cos^220^\circ)$
Using Double angle formula $\cos2A=2\cos^2A-1,$
$\displaystyle N=1-2\cos40^\circ(1+\cos40^\circ)=1-2\cos40^\circ-2\cos^240^\circ=-2\cos40^\circ-(2\cos^240^\circ-1)$
$\displaystyle=-\cos40^\circ-\underbrace{(\cos40^\circ+\cos80^\circ)}$
Using Prosthaphaeresis Formulas on the under-braced part,
$\displaystyle N=-\cos40^\circ-2\cos60^\circ\cos20^\circ=-(\cos40^\circ+\cos20^\circ)$
Again, $\displaystyle\cos40^\circ+\cos20^\circ=2\cos10^\circ\cos30^\circ$
$$\implies\frac{1-4\cos^220^\circ\cos40^\circ}{\sin80^\circ}=-\frac{2\cos10^\circ\cos30^\circ}{\sin80^\circ}=-2\cos30^\circ=-\sqrt3$$

Actually, the problem came into being as
$$\cot20^\circ-\cot30^\circ=\frac{\cos20^\circ}{\sin20^\circ}-\frac{\cos30^\circ}{\sin30^\circ}=\frac{\sin(30^\circ-20^\circ)}{\sin30^\circ\sin20^\circ}$$
$$=\frac{\sin10^\circ}{\frac12\cdot2\sin10^\circ\cos10^\circ}=\frac1{\cos10^\circ}=\frac1{\sin80^\circ}$$
A: This is not an answer but rather a lengthy comment on the connections between your equation and some geometrical problems.
Curiously, your equation can be obtained when trying to solve one of the "famous" $80-20-20$ triangle problems, where $\triangle ABC$ is an isosceles triangle with $AC \cong BC$, $\measuredangle CAB = 80^\circ $, and $D$ on $BC$ such that $CD \cong AB$ (see Figure below). Then letting $x = \measuredangle CDA$, sine rule applied to $\triangle CDA$ and $\triangle DAB$ leads to the equation of OP. There are various purely geometrical approaches to show that $\measuredangle CDB = 150^\circ$, which I think are a very fast way to demonstrate that OP's equation has general solution $$x = 150^\circ + 180^\circ \cdot k, \ \ k \in \Bbb Z,$$
as per the other answers.

If you do not recognize the original version of the problem, you can actually do some sort of reverse-engineering and construct by yourself a very similar one, starting from the equation you have. Here's how.
Consider the Figure below, where again $\triangle ABC$ is isosceles and $\measuredangle CAB = 80^\circ$. For simplicity let $\overline{AC} = 1$. Draw $CH \perp AB$, $PH\perp AC$ and such that $CP \cong CH$, $PQ\perp CH$ and $CR \cong PH$.

Show that

*

*$\overline{CH} = \sin 80^\circ = \overline CR$, and $\overline{AH} = \sin 10^\circ$.

*$\overline{PH} = 2\sin 10^\circ \cos 10^\circ = \sin 20^\circ = \overline{CR}$.

*$\overline{PQ} = \sin 80^\circ \sin 20^\circ$ and $\overline{CQ} = \sin 80^\circ \cos 20^\circ$.

*From the previous point, $$\cot \measuredangle PRQ = \frac{\overline{RQ}}{\overline{PQ}}=\frac{\sin 80^\circ \cos 20^\circ - \sin 20^\circ}{\sin 80^\circ \sin 20^\circ}.$$
So, $x$ in OP's equation is just $\measuredangle PRC$, and you can recognize the original $80-20-20$ triangle problem, on the isosceles triangle $\triangle CPH$.
