Finding smallest root of equation 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:
$-\tan\left(\frac{a}{\sqrt2}\right)=\sqrt2\tan({a/2})$
 A: 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
$$
\tan\frac{a}{2}=0,\quad\sqrt2\quad\text{or}\quad-\sqrt2.
$$
A: 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:
$\tan(2b)=\frac{2\tan(b)}{1-\tan^2(b)}$
$\tan(a)=\tan(2b)=0 \quad or -2\sqrt{2} \quad or \quad 2\sqrt{2}$
$\cdots$
