20. [10] There exist several solutions to the equation
$$
1+\frac{\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x},
$$
where $x$ is expressed in degrees and $0^{\circ}<x<180^{\circ}$. Find the sum of all such solutions.
Answer: $320^{\circ}$
Solution: We first apply sum-to-product and product-to-sum:
$$
\begin{array}{c}
\frac{\sin 4 x+\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x} \\
2 \sin (2.5 x) \cos (1.5 x) \sin (2 x)=\sin (4 x) \sin (3 x)
\end{array}
$$

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),
$$
\begin{array}{c}
\sin (2.5 x)=2 \cos (2 x) \sin (1.5 x) \\
\sin (2.5 x)=\sin (3.5 x)-\sin (0.5 x)
\end{array}
$$

Convert into complex numbers, we get
$$
\begin{array}{c}
\left(x^{3.5}-x^{-3.5}\right)-\left(x^{0.5}-x^{-0.5}\right)=\left(x^{2.5}-x^{-2.5}\right) \\
x^{7}-x^{6}-x^{4}+x^{3}+x-1=0 \\
(x-1)\left(x^{6}-x^{3}+1\right)=0
\end{array}
$$

We recognize the latter expression as $\frac{x^{9}+1}{x^{3}+1}$, giving us
$$
x=0^{\circ}, 20^{\circ}, 100^{\circ}, 140^{\circ}, 220^{\circ}, 260^{\circ}, 340^{\circ} \text {. }
$$

The sum of the solutions is
$$
20^{\circ}+60^{\circ}+100^{\circ}+140^{\circ}=320^{\circ} \text {. }
$$