Find an angle in a geometric figure (given) considering triangles 
Question: In the figure below, AC=AB and AD=BC. Find angle $x$.


My attempt: using a geometric approach, consider the following figure (proportions are not exact). 



*

*Using the notation $AC=AB=b$ and $CD=a$, consider point P, in such a way that $AP=b$ and $\angle PAB=60^o$. 

*Therefore, $\triangle APD\cong \triangle CAB$ (SAS), and $PD=b$. 

*We can also conclude that $\triangle PAB$ is equilateral and $PB=b$.

*Finally, $\triangle DPB$ is isosceles with $\angle DPB=160^o$, so we can conclude that $\angle PBD=10^o$ and $\angle x=10^o$ (as $\angle ABP=60^o$). 
Question: is there a trigonometric approach to the problem? I tried using the sinus law but without success. Any hint or full solution using trigonometric methods or other more straightforward approach will be appreciated.
 A: Let $AB=AC=a$.
Thus, since $$BC=2a\sin50^{\circ},$$ $$DC=AD-AC=BC-AC=2a\sin50^{\circ}-a$$ and $$\measuredangle D=40^{\circ}-x,$$ by law of sines we obtain:
$$\frac{\sin{x}}{DC}=\frac{\sin(40^{\circ}-x)}{BC}$$ or
$$\frac{\sin{x}}{2\sin50^{\circ}-1}=\frac{\sin(40^{\circ}-x)}{\sin50^{\circ}}$$ or
$$\frac{1}{2\sin50^{\circ}-1}=\frac{\sin40^{\circ}\cot{x}-\cos40^{\circ}}{\sin50^{\circ}}$$ or $$\sin40^{\circ}\cot{x}=\frac{2\sin50^{\circ}}{2\sin50^{\circ}-1}+\sin50^{\circ}$$ or
$$\cot{x}=\frac{2\sin50^{\circ}+1}{2\sin50^{\circ}-1}\cot40^{\circ}$$ or
$$\tan{x}=\frac{2\sin50^{\circ}-1}{2\sin50^{\circ}+1}\tan40^{\circ}$$ and since
$$\frac{2\sin50^{\circ}-1}{2\sin50^{\circ}+1}\tan40^{\circ}=\frac{2\cos40^{\circ}-1}{2\cos40^{\circ}+1}\cdot\frac{\sin40^{\circ}}{\cos40^{\circ}}=\frac{\sin80^{\circ}-\sin40^{\circ}}{1+\cos80^{\circ}+\cos40^{\circ}}=$$
$$=\frac{2\sin20^{\circ}\cos60^{\circ}}{1+2\cos20^{\circ}\cos60^{\circ}}=\frac{\sin20^{\circ}}{1+\cos20^{\circ}}=\frac{2\sin10^{\circ}\cos10^{\circ}}{2\cos^210^{\circ}}=\tan10^{\circ},$$
we are done!
A: Just one application of Law of Sines is enough. Let me reproduce the diagram below for convenience.
Denote $\angle ACB = \angle ABC$ as $\theta$, which is given as $\theta = 40^{\circ} = \frac{2\pi}9$.
Note that $\angle CDB = \theta - x$ because $\angle ACB$ is the external angle of $\angle DCB$ for $\triangle BCD$. Apply the Law of Sines on $\triangle BCD$, using your notations for the lengths $a$ and $b$:
$$\frac{ \sin x }a = \frac{ \sin(\theta - x) }{ 2b \cos \theta } \qquad \text{, where the $2b\cos\theta$ comes from} \triangle ABC ~~\text{being isosceles}$$

With the given $|AD| = |BC| \implies a+b = 2b \cos\theta~$, namely, $a = b(2 \cos\theta - 1)$, we have
$$\sin x \,\frac{ 2\cos\theta }{ 2 \cos\theta - 1} = \sin \theta \cos x - \cos\theta \sin x \\
\sin x \,\cos\theta \left( 1+\frac{ 2 }{ 2 \cos\theta - 1} \right)= \sin \theta \cos x \\
\tan x = \tan\theta \frac{ 2 \cos\theta - 1}{ 2 \cos\theta + 1}
$$
Now, looking at this expression, it is tempting to considering something like $x \overset{?}{=} \frac{\theta}4$ by applying twice the half-angle tangent formula. However, this equation holds only for the given $\theta = \frac{2\pi}9$ and one might as well directly consider the analytic properties of $40^{\circ}$ and $10^{\circ}$.
Specifically, for $t\equiv \tan 10^{\circ}$, WolframAlpha tells us that it is a root to the polynomial $p(t)=3t^6 - 27 t^4 + 33 t^2 - 1$ (please scroll down a bit to see that entry).
At the same time, $\displaystyle u \equiv \tan\theta \frac{ 2 \cos\theta - 1}{ 2 \cos\theta + 1}$ is also shown by WolframAlpha to be a root of the same polynomial. Namely $p(u)=0$ just like $p(t) = 0$.
There are six roots (real or complex) of a sextic polynomial. Nonetheless, given our geometric setting (e.g. the known restrictions on the variables) there is no doubt that $$\tan x = u = t = \tan 10^{\circ}$$
Using WolframAlpha is just to out-source the laborious algebraic process. Everything about the tangents being the root of the polynomial are well-established results that can be derived by hand.
