Solve for $x$ in the $80^\circ$-$80^\circ$-$20^\circ$ triangle [closed]

I've been solving this for days but I still I couldn't solve this. Can I know how to solve $x$ ?

I fail to see the importance and difficulty of Langley’s Problem mentioned by Justin Benfield in his answer here apart perhaps some historical mention or maybe some interesting applications. In effect, here I give an EASY generalization of the problem proposed by the OP. I would appreciate any substantiated comment against (am I wrong?).

Let $\triangle{ABC}$ not necessarily isosceles. Given side $a$ and angles $\alpha,\beta,\gamma,\delta$ determine the angles $\phi_1,\phi_2,\phi_3,x,y$ as well as the segments $BE,BD,CD$ and the said difficulty would be to determine $x$ or $y$ since $\phi_1,\phi_2,\phi_3$ are immediate to calculate

We have $$BE=\frac{\sin \beta}{\sin \phi_1}\cdot a\\BD=\frac{\sin \phi_3}{\sin \phi_2}\cdot a\\CD=\frac{\sin \alpha}{\sin \phi_2}\cdot a\\ED=BE\cdot\frac{\sin \delta}{\sin x}=\frac{\sin \beta\sin\delta}{\sin \phi_1\sin x}\cdot a $$ It follows the easy system $$\begin{cases}x+y=\phi_3\\\space\\\dfrac{\sin (\phi_1+\phi_3-x)}{\sin x}=\dfrac{\sin\phi_1\sin\phi_3}{\sin\beta\sin\phi_2}\end{cases}$$ This gives an elementary equation of the form $$A\sin x+B\cos x=C$$ one of the quite popular way to solve is the following: $$A\sin x+B\cos x=C\iff\frac{A\sin x}{\sqrt{A^2+B^2}}+\frac{B\cos x}{\sqrt{A^2+B^2}}=\frac{C}{\sqrt{A^2+B^2}}$$ so $$x+\theta= \arcsin{\frac{C}{\sqrt{A^2+B^2}}}\space\text{where } \space \theta=\arccos{\frac{A}{\sqrt{A^2+B^2}}}$$

Your's is a version of Langley's problem, with slightly different angles. It shouldn't be too difficult to adapt the solution presented here: http://www.gogeometry.com/LangleyProblem.html to your current problem.

Note that this is problem's solution is rather clever, and involves constructing a new edge rather than simple angle chasing (I spent quite a lot of time trying to solve it with angle chasing arguments to no avail before resorting to google searching about it, only to find out that angle chasing didn't cut it, and hence I am now writing this answer).