We have on my Page, at No. 6, the Identity
$\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\frac{1}{\sqrt{3}}$
It is equivalent to $\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\tan30^{\circ}$ or else
$$\tan20^{\circ}\cdot \tan30^{\circ}\cdot \tan40^{\circ}=\tan10^{\circ} \tag{1}$$
Let's consider the equation
$$ \tan x\cdot \tan\frac{3x}{2}\cdot \tan 2x=\tan \frac{x}{2}\tag{2}$$
(1) shows that $x=20^{\circ}$ is a solution for (2).
The problem of completely solving this equation remains open for now.
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