How how do I find the measure of the sides of an equilateral triangle inscribed in a 30-60-90 triangle? 
I am completely lost, I have no idea where to even start. I am sure someone here would be able to do this easily which is why I'm posting this here. Basically the point of the problem is to find the measure of line X. All that is given is that the two segments on the bottom are congruent and equal to 1 unit each which totals to 2 units for the base of the 30-60-90 triangle. If someone could tell me the answer and an explanation on how to get there it would be very much appreciated. 
Just to clarify, the 30s and 60s are degrees, but I made this illustration and I was too lazy to add in the degree symbol. Also please let me know if there is not enough information to solve the problem, but I doubt that is the case because this was a challenge problem that my teacher gave me.


Let the side length of the equilateral triangle be $x$, and let the tilt angle $\theta$ be as shown.  Then
$ x \cos \theta = 1 $
And the top right vertex of the equilateral triangle has coordinates
(Assuming the right angle vertex is the origin)
$P = (1 - x \cos(\theta + 60^\circ) , x \sin(\theta + 60^\circ) ) $
And this point lies on the line whose equation is $y = 2 \sqrt{3} - \sqrt{3} x $
Hence,
$ x \sin(\theta + 60^\circ) = 2 \sqrt{3} - \sqrt{3} (1 - x \cos(\theta+60^\circ)) $
Expanding, and multiplying through by $2$,
$ x ( \sin \theta + \sqrt{3} \cos \theta )  = 2 \sqrt{3} + \sqrt{3} x ( \cos \theta - \sqrt{3} \sin \theta ) $
Simplifying,
$  x \sin \theta =  \dfrac{\sqrt{3}}{2}$
And since $x \cos \theta = 1 $, then $\tan \theta = \dfrac{\sqrt{3}}{2} $
And therefore, $\cos \theta = \dfrac{1}{\sqrt{ 1 + \dfrac{3}{4}}} = \dfrac{2}{\sqrt{7}} $
So that,  $ x =  \dfrac{\sqrt{7}}{2} $
