HMMT February, 2020

Solution:
We first apply sum-to-product and product-to-sum:
$$\frac{\sin 4x + \sin x}{\sin 4x} = \frac{\sin 3x}{\sin 2x}$$
$$2 \sin(2.5x) \cos(1.5x) \sin(2x) = \sin(4x) \sin(3x)$$
Factoring out $\sin(2x) = 0$,
$$\sin(2.5x) \cos(1.5x) = \cos(2x) \sin(3x)$$
Factoring out $\cos(1.5x) = 0$ (which gives us $60^{\circ}$ as a solution),
$$\begin{gathered} \sin(2.5x) = 2 \cos(2x) \sin(1.5x) \\ \sin(2.5x) = \sin(3.5x) - \sin(0.5x) \end{gathered}$$
Convert into complex numbers, we get
$$\begin{gathered} \left(x^{3.5} - x^{-3.5}\right) - \left(x^{0.5} - x^{-0.5}\right) = \left(x^{2.5} - x^{-2.5}\right) \\ x^{7} - x^{6} - x^{4} + x^{3} + x - 1 = 0 \\ (x-1)\left(x^{6} - x^{3} + 1\right) = 0 \end{gathered}$$
We recognize the latter expression as $\frac{x^{9} + 1}{x^{3} + 1}$, giving us
$$x = 0^{\circ},\ 20^{\circ},\ 100^{\circ},\ 140^{\circ},\ 220^{\circ},\ 260^{\circ},\ 340^{\circ}.$$
The sum of the solutions is
$$20^{\circ} + 60^{\circ} + 100^{\circ} + 140^{\circ} = 320^{\circ}.$$