If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$ then quotient $\frac xy$ is equal to? 
If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$, then quotient $\frac xy$ is equal to?

Other conditions are ($\alpha \neq k\pi,\ \alpha \neq \frac \pi2+ k\pi,\ k\in\mathbb Z,\ i^2 = -1$). I tried this by making nominator the difference of squares but it does not lead me anywhere. Solution for this task is $-2-4\cos{2\alpha}$

Lethal weapon #2: Multiply and divide by the same thing:
$$
\frac{1+i\sin{\alpha}}{1-i\sin{3\alpha}}=\frac{i}{i}\frac{1+i\sin{\alpha}}{1-i\sin{3\alpha}}=\frac{i-\sin\alpha}{i+\sin 3\alpha}.
$$
Thus $2x=-\sin\alpha$ and $y=\sin3\alpha$, so
$$
\frac{x}{y}=-\frac{\sin\alpha}{2\sin3\alpha}=-\frac{1}{2+4\cos2\alpha}.
$$
