I considered the more general formula :
$$T(m):=\prod_{k=1}^m \tan\left(\frac{k\pi}{4m}\right)$$
and noticed that the result for small values of $m$ was solution of a polynomial of degree $\le m$.
For $m=45$ I found that the answer was solution of this irreducible polynomial of degree $24$ :
$$1\\- 3256701697315828896312\,x^1 - 325994294876282580655116\,x^2 + 7220097128841103979624568\,x^3 + 112578453555034444841119842\,x^4 + 493898299320136273975435032\,x^5 + 649061666980531722406164708\,x^6 - 840700351973464244018822232\,x^7 -2457988129238279755530778353\,x^8 - 138286882106888055215208624\,x^9 +2474525072938192662606171624\,x^{10}+326024084648343835216068912\,x^{11} - 1088043811994145989051965476\,x^{12} + 5147738954805237173669808\,x^{13} +182273284200850360076819304\,x^{14} - 33045263177263307887100976\,x^{15} + 677463542076505961377071\,x^{16} +170537100491574073221480\,x^{17} - 6714674580553776884700\,x^{18} - 128584156182235814952\,x^{19} - 339010000890501150\,x^{20}\\ +776030507612856\,x^{21} - 397610115660\,x^{22}\\ + 37004040\,x^{23} + 6561\,x^{24}$$
This is an 'experimental' result (no proof) but rather satisfying from the 'not nice' point of view ! ;-)
Rewrite $\tan x= \tan2x\tan3x\tan4x$ as
$$\sin x \cos 2x\cos3x\cos 4x = \cos x \sin 2x\sin 3x\sin 4x$$
and factorize,
$$\sin x\cos 2x (\cos3x\cos 4x -4\cos^2 x\sin 3x\sin 2x)=0$$
Further factorize with $\cos 3x = \cos x(2\cos 2x -1)$,
$$\sin x\cos 2x \cos x [(2\cos 2x -1)\cos 4x -4\cos x\sin 3x\sin 2x)]=0\tag 1$$
Recognize $\cos x \ne 0$, $\cos 2x \ne 0$ and
$$(2\cos 2x -1)\cos 4x =\cos2x+\cos6x-\cos4x$$
$$4\cos x\sin 3x\sin 2x=2(\sin4x+\sin2x)\sin2x= \cos2x-\cos6x+1-\cos4x$$
to reduce the equation (1) to,
$$\sin x(2\cos 6x -1)=0 $$
which leads to $\sin x =0$ and $\cos6x=\frac12$. Thus, the solutions are
$$x=n\pi,\>\>\>\>\> x = \frac{n\pi}3\pm\frac\pi{18}$$
Best Answer
$$\tan x=\tan 2x\tan 4x\tan 8x$$
$$\frac{\sin x}{\cos x}=\frac{2 \sin x \cos x}{\cos 2x}\left(\frac{2\sin 4x \sin 8x}{2 \cos 4x \cos 8x}\right) $$
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