<s>[INST] The equation $\tan x = \tan 2x \tan 4x \tan 8x$ In the question  we have the equality
$$\tan 6^{\circ} \tan 42^{\circ} = \tan 12^{\circ} \tan 24^{\circ}$$
which is equivalent to
$$ \tan 6^{\circ} = \tan 12^{\circ} \tan 24^{\circ} \tan 48^{\circ}$$
This means that the equation
$$\tan x = \tan 2x \tan 4x \tan 8x$$
has the solution $x =6^{\circ} = \frac{\pi}{30}$.  How to find all the solution of this equation?
 [/INST] Factorize the equation as follows
\begin{align}
& \tan x-\tan 2x \tan 4x \tan 8x \\
=& \frac{\sin x \cos 2x \cos 4x \cos 8x-\cos x \sin 2x \sin 4x \sin 8x }
{\cos x \cos 2x \cos 4x \cos 8x }\\
=& \frac{\sin x [\cos 2x \cos 4x \cos 8x-(1+\cos 2x)\sin 4x \sin 8x ]}
{\cos x \cos 2x \cos 4x \cos 8x }\\
 =& \frac{\tan x (\cos 12x -2\sin 2x \sin 8x )}
{\cos 4x \cos 8x }\\
=& \frac{\tan x }{\cos8x} \left( 4 \cos^2 4x-3 -4\sin 2x \sin 4x\right)\\
 =& \frac{\tan x }{\cos8x} \left(16\cos^4 2x +8\cos^3 2x -16\cos^2 2x-8\cos 2x +1 \right)\\
 =& \frac{16\tan x }{\cos8x} \left(\cos^2(2x-\frac{2\pi}3) - \frac12 \cos(2x-\frac{2\pi}3) -\frac14\right)\\
&\hspace{15mm}\times \left(\cos^2(2x+\frac{2\pi}3) - \frac12 \cos(2x+\frac{2\pi}3) -\frac14\right)\\
 =& \frac{16\tan x }{\cos8x} \left(\cos(2x-\frac{2\pi}3) - \cos\frac\pi5\right)\left(\cos(2x-\frac{2\pi}3) - \cos\frac{3\pi}5\right)\\
&\hspace{15mm}\times \left(\cos(2x+\frac{2\pi}3) - \cos\frac\pi5\right)\left(\cos(2x+\frac{2\pi}3) - \cos\frac{3\pi}5\right)\\
\end{align}
where $ \cos\frac{\pi}5 +\cos\frac{3\pi}5 =\frac12$ and $ \cos\frac{\pi}5\cos\frac{3\pi}5 =-\frac14$ are recognized. Then, the full set of the solutions are $x=n\pi,\>n\pi \pm \frac{\pi}3\pm \frac\pi{10}$ and $n\pi \pm \frac{\pi}3\pm \frac{3\pi}{10}
$, or
$$x-n\pi=0, \>\pm \frac\pi{30}, \>\pm \frac{7\pi}{30} , \>\pm \frac{13\pi}{30} , \>\pm \frac{19\pi}{30}
$$
</s>