How find the equation $\cot x=\frac{\sin 20^\circ - \sin 80^\circ \cos 20^\circ}{\sin 80^\circ \sin 20^\circ}$

Question

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

Answer 1

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 $x\in\left\{-\frac{\pi}{6}+\pi k:k\in\Bbb{Z}\right\}$.$\qquad\square$

Answer 2

$$\sin80^\circ=2\sin40^\circ\cos40^\circ=2(2\sin20^\circ\cos20^\circ)\cos40^\circ$$

$$\implies\csc80^\circ-\cot20^\circ=\frac{1-4\cos^220^\circ\cos40^\circ}{\sin80^\circ}$$

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, $\displaystyle\cos40^\circ+\cos20^\circ=2\cos10^\circ\cos30^\circ$

$$\implies\frac{1-4\cos^220^\circ\cos40^\circ}{\sin80^\circ}=-\frac{2\cos10^\circ\cos30^\circ}{\sin80^\circ}=-2\cos30^\circ=-\sqrt3$$

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Actually, the problem came into being as

$$\cot20^\circ-\cot30^\circ=\frac{\cos20^\circ}{\sin20^\circ}-\frac{\cos30^\circ}{\sin30^\circ}=\frac{\sin(30^\circ-20^\circ)}{\sin30^\circ\sin20^\circ}$$

$$=\frac{\sin10^\circ}{\frac12\cdot2\sin10^\circ\cos10^\circ}=\frac1{\cos10^\circ}=\frac1{\sin80^\circ}$$

Answer 3

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

1. $\overline{CH} = \sin 80^\circ = \overline CR$, and $\overline{AH} = \sin 10^\circ$.
2. $\overline{PH} = 2\sin 10^\circ \cos 10^\circ = \sin 20^\circ = \overline{CR}$.
3. $\overline{PQ} = \sin 80^\circ \sin 20^\circ$ and $\overline{CQ} = \sin 80^\circ \cos 20^\circ$.
4. 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 $\triangle CPH$.