We first apply sum-to-product and product-to-sum: $\frac{\sin 4 x+\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x}$. $2 \sin (2.5 x) \cos (1.5 x) \sin (2 x)=\sin (4 x) \sin (3 x)$. Factoring out $\sin (2 x)=0$, $\sin (2.5 x) \cos (1.5 x)=\cos (2 x) \sin (3 x)$. Factoring out $\cos (1.5 x)=0$ (which gives us $60^{\circ}$ as a solution), $\sin (2.5 x)=2 \cos (2 x) \sin (1.5 x)$. Convert into complex numbers, we get $\left(x^{3.5}-x^{-3.5}\right)-\left(x^{0.5}-x^{-0.5}\right)=\left(x^{2.5}-x^{-2.5}\right)$. $x^{7}-x^{6}-x^{4}+x^{3}+x-1=0$. $(x-1)\left(x^{6}-x^{3}+1\right)=0$. 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 $20^{\circ}+60^{\circ}+100^{\circ}+140^{\circ}=320^{\circ}$.