find the angle of X

Applying the Sine Rule in a variety of triangles, we have $\frac{AB}{AD} \; = \; \frac{\sin 3x}{\sin x} \hspace{1cm} \frac{AD}{AC} \; = \; \frac{\sin3x}{\sin8x} \hspace{1cm} \frac{AC}{AB} \; = \; \frac{\sin14x}{\sin7x} \; = \; 2\cos7x$ and hence we deduce that $\frac{2\sin^23x \cos7x}{\sin x \sin 8x} \; = \; 1$ and hence that $2\sin^23x \cos7x \; = \; \sin x \sin 8x$ and we must solve this equation with $x > 0$ and $14x < 180^\circ$ . It turns out that $x=\boxed{10^\circ}$ is the only solution in this range.

Sorry, but I think that every x< $\frac{90}{7}$ is a possible answer to this problem. I guess that the answer was 10 because a lot of people love to set problems with rounded numbers... But, 10 isn't the only possible solution.