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I have the following system of trig equation. I was wondering if this system has a solution or not. Mathematica doesn't solve it but I think it is very easy to solve unless I am missing something.

It is intended to solve it for L.

$\alpha_1 = \alpha_2\ Tan(\frac{-\omega\ L}{2})$

$\alpha_1 = \alpha_2\ Tan(\frac{\omega\ L}{2})$

What normally I do is to put:

$\alpha_2\ Tan(\frac{-\omega\ L}{2}) = - \alpha_2\ Tan(\frac{\omega\ L}{2})$

And then I will have two equations:

$\alpha_1 = -\alpha_2\ Tan(\frac{\omega\ L}{2})$

$\alpha_1 = \alpha_2\ Tan(\frac{\omega\ L}{2})$

And then:

$-\alpha_2\ Tan(\frac{\omega\ L}{2}) = \alpha_2\ Tan(\frac{\omega\ L}{2}) $

$2\ \alpha_2\ Tan(\frac{\omega\ L}{2}) =0 $

And since my $\alpha_2$ is not zero, as $\alpha_1$ wasn't, therefore:

$2\ Tan(\frac{\omega\ L}{2}) =0 $

Consequently:

$Tan(\frac{\omega\ L}{2}) =0 $

$\frac{\omega\ L}{2} = n \pi $

If $\omega$ is a number, then:

$L = \frac{2\ n\ \pi}{\omega}$

If the solution is correct, why Mathematica doesn't solve it? If it is not, so...

Thanks in advance

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  • $\begingroup$ No I didn't. What kind? $\endgroup$ – KratosMath Dec 6 '16 at 18:49
  • $\begingroup$ any condition for $\alpha_i$? $\endgroup$ – Arnaldo Dec 6 '16 at 18:49
  • $\begingroup$ Yes, All the variables are greater than zero. $\endgroup$ – KratosMath Dec 6 '16 at 18:50
  • $\begingroup$ No condition for $\alpha_i$ $\endgroup$ – KratosMath Dec 6 '16 at 18:50
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If $\alpha_2=0$ then $\alpha_1=0$ and $L$ can be anyone.

If $\alpha_2 \ne 0$ and $\alpha_1=0$ then $\tan(-wL/2)=\tan(wL/2)=0 \Rightarrow wL/2=k\pi \Rightarrow L=2k\pi/w$ if $w \ne 0$, if not, $L$ can be anyone.

If $\alpha_1.\alpha_2 \ne 0$ then $\alpha_1/\alpha_2=\tan(-wL/2)=\tan(wL/2)$ but $\tan(x)$ is a odd function so, $2\tan(wL/2)=0 \Rightarrow wL/2=k\pi \Rightarrow L=2k\pi/w$

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  • $\begingroup$ Exactly. But why from the begining mathematica doesn't solve? $\endgroup$ – KratosMath Dec 6 '16 at 19:00
  • $\begingroup$ I don't know exactly, maybe you didn't give all informations like constrains for $\alpha$ but that is just a guess. $\endgroup$ – Arnaldo Dec 6 '16 at 19:02
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solve the equation $$\tan(\omega L/2)=\tan(-\omega L/2)$$ per hand

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  • $\begingroup$ So you mean that, My hand calculation is correct. right? $\endgroup$ – KratosMath Dec 6 '16 at 18:59

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