A forgotten trigonometric equation // Egy elfeledett trigonometrikus egyenlet

 We will solve the following trigonometric equation here :

$$\tan(30^{\circ}-x)\cdot \tan(30^{\circ}+x)=\frac{\tan x}{\sqrt{3}} \tag{1}$$

I first mentioned it in the Post here. There we also showed that the equation  (1)  is equivalent to the equation

$$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0,\;\;t=\tan x \tag{2}$$

ANSWER CiP

$$x_k^{\circ}=20^{\circ}+k\cdot 60^{\circ},\;\;k\in\mathbb{Z} \tag{3}$$

                    Solution CiP

              The calculation has already been done, but we are presenting it for convenience.

$(1)\overset{t=\tan x}{\Leftrightarrow} t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0\Leftrightarrow \sqrt{3}=\frac{3t-t^3}{1-3t^2}\Leftrightarrow$

$\Leftrightarrow \sqrt{3}=\frac{3\cdot \tan x-\tan^3x}{1-3\cdot \tan^2x}\Leftrightarrow \tan 60^{\circ}=\tan 3x\Leftrightarrow 3x=60^{\circ}+k\cdot 180^{\circ},\;\;k\in \mathbb{Z}$

From here we get the answer  (3)

$\blacksquare$

          Remark CiP   It all started with solving the first equation from No. 6 on Page here. Now, after we have solved the equation  (1) (see also the Post from September 3, 2025), we can write the identities :

$$\frac{\tan10^{\circ}\cdot \tan50^{\circ}}{\tan20^{\circ}}=\frac{1}{\sqrt{3}}\tag{4}$$

$$\frac{\tan10^{\circ}\cdot \tan70^{\circ}}{\tan40^{\circ}}=\frac{1}{\sqrt{3}}\tag{5}$$

$$\frac{\tan50^{\circ}\cdot \tan70^{\circ}}{\tan80^{\circ}}=\frac{1}{\sqrt{3}}\tag{6}$$

          But  (4), considering that  $\frac{1}{\tan20^{\circ}}=\cot20^{\circ}$, is exactly the first identity from No. 6. Seeking to express all quantities as tangents of angles not exceeding  $45^{\circ}$, we further have  $\frac{\tan10^{\circ}\cdot \tan50^{\circ}}{\tan20^{\circ}}=\frac{\tan10^{\circ} \cdot \color{Red}{\cot40^{\circ}}}{\tan20^{\circ}}=\frac{\tan10^{\circ}\cdot \color{Red}{\frac{1}{\tan40^{\circ}}}}{\tan20^{\circ}}=\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\frac{1}{\sqrt{3}}$

which is the second entity from the aforementioned No. 6.

     Doing the same thing with  (5) : 

$\frac{1}{\sqrt{3}}=\frac{\tan10^{\circ}\cdot \color{Red}{\cot20^{\circ}}}{\tan40^{\circ}}=\frac{\tan10^{\circ}\cdot \color{Red}{\frac{1}{\tan20^{\circ}}}}{\tan40^{\circ}}=\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan{40^{\circ}}}$

and with (6) :

$\frac{1}{\sqrt{3}}=\frac{\color{Red}{\cot40^{\circ}\cdot \cot20^{\circ}}}{\color{Red}{\cot10^{\circ}}}=\frac{\frac{1}{\tan40^{\circ}}\cdot \frac{1}{\tan20^{\circ}}}{\frac{1}{\tan10^{\circ}}}=\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}$

So the second of the identities in No. 6 is the most important.

              Let's demonstrate it directly then. We will see that the pattern is the same.

$\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\frac{\frac{\sin10^{\circ}}{\cos10^{\circ}}}{\frac{\sin20^{\circ}}{\cos20^{\circ}}\cdot \frac{\sin40^{\circ}}{\cos40^{\circ}}}=\frac{8\cdot \sin10^{\circ}\cdot \cos20^{\circ}\cdot \cos40^{\circ}}{8\cdot \cos10^{\circ}\cdot \sin20^{\circ}\cdot \sin40^{\circ}}\tag{7}$

The numerator in (7) is  $4\cdot \frac{\sin20^{\circ}}{\cos10^{\circ}}\cdot \cos20^{\circ}\cdot \cos40^{\circ}=\frac{1}{\cos10^{\circ}}\cdot 2\cdot \sin40^{\circ}\cdot \cos40^{\circ}=\frac{1}{\cos10^{\circ}}\cdot \sin80^{\circ}=1$

The denominator in (7) is  $4\cdot \cos10^{\circ}\cdot(\cos20^{\circ}-\cos60^{\circ})=4\cos10^{\circ}\cdot \cos20^{\circ}-4\cos10^{\circ}\cdot \frac{1}{2}=$

$=2(\cos10^{\circ}+\cos30^{\circ})-2\cos10^{\circ}=2\cos10^{\circ}+2\cdot \frac{\sqrt{3}}{2}-2\cos10^{\circ}=\sqrt{3}$

So the final value in (7) is  $\frac{1}{\sqrt{3}}$.

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