<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] $$\tan x=\tan 2x\tan 4x\tan 8x$$
$$\frac{\sin x}{\cos x}=\frac{2 \sin x \cos x}{\cos 2x}\left(\frac{2\sin 4x \sin 8x}{2 \cos 4x \cos 8x}\right) $$
If $\sin x\neq0$ then,
$$\frac{\cos 2x}{2\cos^2 x}=\frac{\cos 4x - \cos 12x}{\cos 4x + \cos 12x}$$
Now,using componendo-dividendo,
$$\frac{\cos 2x+ 2\cos^2x}{2\cos^2 x-\cos 2x}=\frac{\cos 4x}{\cos 12x}$$
$$ 4 \cos^2 x -1 =\frac{1}{4\cos ^2 4x-3}$$
$$(2\cos 2x+1)(4(2\cos^2 2x -1)^2-3)=1$$
These are the solutions of above equation :wolframalpha
can someone solve the last equation by hand and edit the answer$?$
also,note that I cancelled some terms like $\cos 4x$,because $\cos 4x=0$ is not in the domain of the equation
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