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
$$11 \tan x = 6 + \sqrt3$$
 A: Simple: isolate $\tan x$ in one side. You get:
$$\frac{2\sqrt3-1}{2}\,\tan x=\frac{\sqrt3}2 \enspace\text{whence}\enspace \tan x=\frac{\sqrt3}{2\sqrt3-1}$$
Then rationalise the denominator:
$$\tan x=\frac{\sqrt3(2\sqrt3+1)}{12-1}=\frac{6+\sqrt3}{11}.$$
A: I will try to make it a bit more elaborate since you are going to need to be able to manipulate equations like this all the time. 
Multiply both sides by 2:
$\implies \sqrt{3}+\tan(x)=2\tan(x)\sqrt{3}$
Subtract $\tan(x)$ from both sides in order isolate $\tan(x)$
$\implies \sqrt{3}=2\tan(x)\sqrt{3}-\tan(x)$ 
You have two terms with $\tan(x)$ on the RHS so you can factor out $\tan(x)$
$\implies \sqrt{3}=\tan(x)(2\sqrt{3}-1)$
Divide both sides by $(2\sqrt{3}-1)$
$\implies \tan(x)=\frac{\sqrt{3}}{2\sqrt{3}-1}$
Now you can multiply the fraction on the RHS by $\frac{2\sqrt{3}+1}{2\sqrt{3}+1}$ in order to get rid of the sqare root in the denominator. Remember you are just multiplying by 1 so it doesn't change the equation
$\implies \tan(x)=\frac{\sqrt{3}}{2\sqrt{3}-1}\cdot \frac{2\sqrt{3}+1}{2\sqrt{3}+1}$
Just do a bit of multiplication and you will get
$\tan(x)=\frac{6+\sqrt{3}}{11} \iff 11\tan(x)=6+\sqrt{3} $
A: $$\frac{\sqrt{3}}{2\sqrt{3}-1}=\tan(x)$$
$$\frac{\sqrt{3}}{2\sqrt{3}-1}*\frac{2\sqrt{3}+1}{2\sqrt{3}+1}=\tan(x)$$
$$\frac{6+\sqrt{3}}{11}=\tan(x)$$
