Simple algebra involving trigonometry, but confusing How do I get from 
$$\frac{\sqrt3}{2} + \frac12 \tan x = 2 \tan x \cdot \frac{\sqrt3}{2}$$
to
$$\frac{\sqrt3}{2 \sqrt3 - 1} = \tan x$$
and then to
$$11 \tan x = 6 + \sqrt3$$

$$\frac{\sqrt{3}}{2\sqrt{3}-1}=\tan(x)$$
$$\frac{\sqrt{3}}{2\sqrt{3}-1}*\frac{2\sqrt{3}+1}{2\sqrt{3}+1}=\tan(x)$$
$$\frac{6+\sqrt{3}}{11}=\tan(x)$$
