My friend asked me a geometric problem.
In $\triangle ABC$, $\angle B=\angle C=70^\circ$. $D$ is an interior point of the triangle such that $\angle BCD=40^\circ$ and $\angle CBD=20^\circ$. Find $\angle BAD$.
If $\angle BAD=\theta$, it is not difficult to see that $\dfrac{\tan(40^\circ-\theta)}{\tan40^\circ}=\dfrac{\tan30^\circ}{\tan70^\circ}$ and I managed to show that $\tan(40^\circ-\theta)=\tan 10^\circ$ using trigonometric identities.
I actually proved that $\tan10^\circ\tan70^\circ=\tan30^\circ\tan40^\circ$, or equivalently, $\tan10^\circ=\tan20^\circ\tan30^\circ\tan40^\circ$. This result is so beautiful and make me interested in the equation $\tan x= \tan2x\tan3x\tan4x$, but I have difficulty in solving it. By plotting the graph, I can see that the solution is $180n^\circ$ or $60n^\circ\pm10^\circ$.
My questions are
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How to solve the original geometric problem without using trigonometry?
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How to solve the equation $\tan x= \tan2x\tan3x\tan4x$?
Remark
Just find a solution to the second question a few minutes after posting it. But I still want to see if there are other ways to solve it.
Best Answer
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
$$x=n\pi,\>\>\>\>\> x = \frac{n\pi}3\pm\frac\pi{18}$$