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?

Another solution.
Consider the equation
$$\tan (2x) \tan (4x) \tan (8x)-\tan(x)=0$$ Let $x=\tan^{-1}(t)$, use the multiple angle formulae and obtain
$$\frac{t^9-28 t^7+134 t^5-92 t^3+t}{t^8-28 t^6+70 t^4-28 t^2+1}=0$$ Discrading the trivial $t=0$,we then face a quartic equation in $u=t^2$
$$u^4-28 u^3+134 u^2-92 u+1=0$$ which can be solved with radicals. The solutions are all positive and the smallest is
$$u=7-2 \sqrt{5}-2 \sqrt{15-6 \sqrt{5}}\implies t=\sqrt{7-2 \sqrt{5}-2 \sqrt{15-6 \sqrt{5}} }$$ and then
$$x=\tan ^{-1}\left(\sqrt{7-2 \sqrt{5}-2 \sqrt{15-6 \sqrt{5}}}\right)$$ At this point, I am stuck; using my 50+ years old pocket calculator
$x=0.1047197551$ which is $\frac \pi{30}$.
But, as @Thomas Andrews answered, there are many other solutions.
