Angles in Isosceles Triangles Find the value of $x$ in the triangle.

My attempt,
I do know that because it is an isosceles triangle, so that $\angle ABC=\angle ACB$
Of course we can see that $y=3x$ and $z=5x$ so that $x=9$ by adding $12x+3x+5x=180$
But how do we know that that's the unique solution?
Basically we can see that $$5x+y=3x+z$$
But how to know $y=3x$ and $z=5x$ are the only possible solution? Thanks in advance.
 A: 
In the figure, $E$ is a point on $CD$ produced such that $\angle CAE=5x$.
$\because  \angle ACE=\angle BAD=3x$, $\angle EAC=\angle DBA=5x$ and $AC=AB$
$\therefore \Delta ACE \cong \Delta BAD $ (ASA)
$\therefore BD=AE$  ----- (1)
$\because \angle EAD=\angle EDA=4x$
$\therefore AE=DE$ ----- (2)
From (1), (2), we have $DE=BD$ and hence $\angle DEB=\angle DBE$
Since $\angle DEB+\angle DBE=12x$
$\therefore \angle DEB=\angle DBE=6x$
This implies that $\angle EBA=6x-5x=x$
$\because \angle EBA=\angle EAB=x$
$\therefore EA=EB$ ----- (3)
(1), (2) and (3) implies that $EB=BD=DE$
$\Delta BED$ is an equilateral triangle.
Hence $\angle BED=60^o$
$6x=60^o$
$x=10^o$
A: It seems this problem is more of a trigonometric equation than a geometry problem. However, by the law of sine, we conclude:
$$\frac {\sin 5x}{\sin 8x}=\frac{AD}{AB}=\frac{AD}{AC}=\frac {\sin 3x}{\sin \angle ADC}=\frac {\sin 3x}{\sin 4x}\implies \sin 8x \sin 3x=\sin 5x \sin 4x$$
$$\implies 2\cos 4x\sin 3x=\sin 5x\implies -\sin x+\sin 7x=\sin 5x$$
$$-\sin x=\sin 5x-\sin 7x=-2\cos 6x\sin x\implies 2\cos 6x=1 \implies x=10^{\circ}.$$
