What's the measure of angle $x$ in the triangle below? 
Any point $D$, which is interior to the triangle $\triangle ABC$, determines vertex angles
having the following measures:
$m(BAD) = x,$
$m(ABD) = 2x,$
$m(BCD) = 3x,$
$m(ACD)= 4x,$
$m(DBC) = 5x.\\$
Find the measure of $x$.
(Answer:$10^\circ$)


My progress:
$\dfrac{AB}{AD}=\dfrac{\sin(180−3x)}{\sin(2x)}\\
\dfrac{AC}{AD}=\dfrac{\sin(11x)}{\sin(4x)}
AB=AC \implies 
\dfrac{\sin(3x)}{\sin(2x)}=\dfrac{\sin(11x)}{\sin(4x)}$
but it is a complex equation to solve. Is there another way?

$\angle(CDB)=5x<\pi\implies x<\dfrac{\pi}{5} $
$\therefore 0<x<\dfrac{\pi}{5}\\
\dfrac{\sin(3x)}{\sin(2x)}=\dfrac{\sin(11x)}{\sin(4x)}\\
\dfrac{\sin(3x)}{ {\sin(2x)}}=\dfrac{\sin(11x)}{2{\sin(2x)}\cos(2x)}\\
2\sin(3x)\cos(2x)=\sin(11x)\\
\sin(5x)+\sin(x)=\sin(11x)\\
\sin(5x)=\sin(11x)-\sin(x)\\
{\sin(5x)}=2\cos(6x){\sin(5x)}\\
\cos(6x)=\dfrac{1}{2}\\
6x=\dfrac{\pi}{3}\\
x=\dfrac{\pi}{18}$
