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 $-2-4\cos{2\alpha}$
 A: $$2x+\sin3a+i(1-2x\sin3a)=y-\sin a+i(1+y\sin a)$$
Eauting the imaginary parts, $1-2x\sin3a=1+y\sin a\iff \dfrac xy=\dfrac{\sin a}{-2\sin3a}$
Now $\sin3a=\sin a(3-4\sin^2a)$ and $\cos2a=1-2\sin^2a\iff2\sin^2a=1-\cos2a$
A: $$(2x+i)(1-i\sin 3\alpha)=(y+i)(1+i\sin \alpha) \Rightarrow (2x+\sin 3\alpha)+i(1-2x\sin 3\alpha) = (y-\sin\alpha)+i(1+y\sin \alpha)$$
Equate the imaginary parts,
$$\Rightarrow \frac{x}{y}=-\frac{\sin \alpha}{2\sin 3\alpha}=-\frac{\sin \alpha}{2(\sin \alpha \cos 2\alpha+\cos \alpha \sin 2\alpha)}$$
Use $\sin2\alpha =2\sin\alpha\cos\alpha$ and $\cos 2\alpha=2\cos^2\alpha-1$
$$\Rightarrow \frac{x}{y}=-\frac{1}{2(\cos 2\alpha +2\cos^2\alpha)}=-\frac{1}{2(2\cos 2\alpha+1)}$$
A: 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
$$
\frac{x}{y}=-\frac{\sin\alpha}{2\sin3\alpha}=-\frac{1}{2+4\cos2\alpha}.
$$
