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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 $\frac{\pi}{4}$

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What does tg5x mean ? –  Amr Nov 24 '12 at 15:28
    
is tg= trigonometric? –  clark Nov 24 '12 at 15:29
    
Corrected. It means tan5x –  Adam Nov 24 '12 at 15:34
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Wolfram Alpha thinks there are more solutions (none of which --apart from the multiplies of $\pi$-- are very attractive): wolframalpha.com/input/… –  Blue Nov 24 '12 at 15:51
    
@Blue: do you mind making your comment as an answer? –  Willie Wong Jan 29 '13 at 13:13
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2 Answers

Multiply $$ \tan(5x)-2\tan(3x)=\tan(3x)\tan(5x) $$ by $\cos(3x)\cos(5x)$ and cancel to get $$ \begin{align} 0 &=\sin(5x)\cos(3x)-2\sin(3x)\cos(5x)-\sin(3x)\sin(5x)\\ &=\tfrac12(\sin(8x)+\sin(2x))-2\cdot\tfrac12(\sin(8x)-\sin(2x))-\tfrac12(\cos(2z)-\cos(8x))\\ &=\tfrac12(\cos(8x)-\sin(8x)+3\sin(2x)-\cos(2x))\\ &=\tfrac12(\sqrt2\sin(8x+\tfrac34\pi)+\sqrt{10}\sin(2x-\arctan(\tfrac13)))\\ \end{align} $$ There doesn't appear to be any simple way to solve this other than numerically.

The function has a period of $\pi$. Plotting, it appears that this function has $4$ roots in $[0,\pi)$.

$\hspace{8mm}$enter image description here

Numerically, the roots in $[0,\pi)$ are $$ x=\left\{\begin{array}{} 0.0000000000000000000\\ 0.3598476319540602998\\ 1.6915164344292107407\\ 3.0829964597422711815 \end{array}\right. $$

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Making my comment into an answer, as requested:

Wolfram Alpha thinks there are more solutions (none of which ---apart from the multiples of $\pi$--- are very attractive). See here.

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