I have been trying to solve this equation for over a week now: $\tan5x-2\tan3x=\tan3x\tan5x$
I found one solution $x=k\pi$ but I cannot prove that this is the only solution. It is equivalent to: $\sin 5x \cos 3x - 2 \sin 3x \cos 5x = \sin 3x \sin 5x$ $\sin 2x =\sqrt 2 \sin 3x \sin(5x+ \frac{\pi}{4})$ $2\sin x \cos x =2\sqrt 2 \sin x \cos (2x +1) \sin(5x+ \frac{\pi}{4})$ So one solution is $x=k\pi$. But how to proceed with the remaining: $\sqrt 2 \cos x = 2\cos (2x +1) \sin(5x+ \frac{\pi}{4}) $ It seems impossible to use derivatives to prove that it has no solution because the argument of the third cosine is shifted by $\frac{\pi}{4}$