Solve inverse trigonometric equation $\frac{\pi}{6}=\tan^{-1} \frac{11}{x} -\tan^{-1} \frac{1}{x}$ How do I go about solving for $x$ when I have:
$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$. 

The formula for the tangent of a difference says
$$
\tan(a-b) = \frac{\tan a-\tan b}{1+\tan a \tan b}.
$$
If $p = \tan a$ and $q=\tan b$ and $a = \tan^{-1}p$ and $b=\tan^{-1} q$ then we have
$$
\tan(a-b) = \frac{p-q}{1+pq}
$$
so
$$
a-b = \tan^{-1} \frac{p-q}{1+pq}
$$
or in other words
$$
\tan^{-1} p - \tan^{-1} q = \tan^{-1}\frac{p-q}{1+pq}.
$$
So plug in the numbers you specified:
$$
\tan^{-1} \frac {11} x - \tan^{-1} \frac 1 x = \tan^{-1} \frac{\frac{11}x - \frac 1 x}{1 + \left(\frac{11} x \cdot \frac 1 x\right)}
$$
$$
= \tan^{-1} \frac{10x}{x^2+11}
$$
Thus
$$
30^\circ = \tan^{-1} \frac{10x}{x^2+11}
$$
$$
\frac{10x}{x^2+11} = \tan 30^\circ = \frac{\sqrt 3}3
$$
$$
30x = \sqrt 3\, x^2 + 11\sqrt 3.
$$
This is a quadratic equation.
