Geometry: Find angle x in triangle I have not been able to find a euclidean geometry solution to this, but any other solutions are also appreciated.

Let ABC be a triangle with AB=CD and angles as marked in the diagram. Find the measure in degrees of angle $x$.
Answer: $15^o$.

let $AB = DC = 1, AD = a.$  the by the rule sine applied to the triangle $ABD,$ we have $$\frac a{\sin x} = \frac 1{\sin 3x}\to a = \frac{\sin x}{\sin 3x} = \frac{\sin x}{\sin x \cos 2x + \sin 2x \cos x} = \frac{1}{\cos 2x + 2\cos^2x} $$
apply the rule of sine to the triangle $ABC,$ then $$\frac{1+a}{\sin 7x} =\frac1{\sin 9x}$$  so we need to solve $$\sin 9x =(\sin 7x - \sin 9x)(\cos 2x + 2\cos^2 x),\, 0 < x < 20^\circ.$$ 
this has one positive solution in the $x = 15^\circ$ in the interval $0 < x < 20^\circ.$
