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By trial and modification, I want to find the smallest root of this equation: $-2\tan\left(\frac{a}{2}\right)=\tan(a)$

and the following as well: $-\tan\left(\frac{a}{\sqrt2}\right)=\sqrt2\tan({a/2})$

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$a = -2n\pi$ is a root for all $n \in \mathbb N$. So, no smallest root. –  TonyK Oct 30 '11 at 13:10
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2 Answers

up vote 0 down vote accepted

Hint:

If $\frac{a}{2}=b$, then you equation became:

$-2\tan(b)-\tan(2b)=0$

But,

$\tan(2b)=\frac{2\tan(b)}{1-\tan^2(b)}$

so we have:

$-2\tan(b)-\frac{2\tan(b)}{1-\tan^2(b)}=0 \iff$

$\frac{-2tan(b)(1-tan^2(b))}{1-tan^2(b)}-\frac{2tan(b)}{1-tan^2(b)}=0$

The denominator couldn't be 0, so:

$1-\tan^2(b) \neq 0 \iff \tan^2(b) \neq 1 \iff (\tan(b) \neq -1 \quad or \quad \tan(b) \neq -1)$

Then the equation become:

$\tan^3(b)=2\tan(b) \iff \tan(b)=0 \quad or \quad \tan(b)=-\sqrt{2} \quad or \quad \tan(b)=\sqrt{2}$

But $a=2b$, so we get:

$\tan(2b)=\frac{2\tan(b)}{1-\tan^2(b)}$

$\tan(a)=\tan(2b)=0 \quad or -2\sqrt{2} \quad or \quad 2\sqrt{2}$ $\cdots$

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Are you searching about the smallest real root in $[-2\pi,2\pi]$ interval? Please reformulate your question... –  GarouDan Oct 30 '11 at 13:22
    
What about -tan(a/sqrt2)-sqrt2*tan(a/2)=0 ? –  Chris Oct 30 '11 at 14:03
    
@Chris. Interesting question. I'm not sure if it solves, but you can try: $\tan(x)=\frac{e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}$ and replace with your values. If you can't solve post here what you got or ask as another question. –  GarouDan Oct 30 '11 at 14:26
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First of all, observe that the function $-2\tan(a/2)-\tan a$ is periodic of period $2\,\pi$. This implies that if $a$ is a root of your equation, so is $a+2\,k\,\pi$ for any integer $k$. In particular, there is no smallest root.

Next, why trial and modification when you can solve it (almost) exactly? Using the formula $$ \tan a =\tan 2(\frac{a}{2})=\frac{2\tan\frac{a}{2}}{1-\tan^2\frac{a}{2}} $$ it is easy to get $$ \tan\frac{a}{2}=0,\quad\sqrt2\quad\text{or}\quad-\sqrt2. $$

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