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 $-2-4\cos{2\alpha}$

$$2x+\sin3a+i(1-2x\sin3a)=y-\sin a+i(1+y\sin a)$$
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 $\cos2a=1-2\sin^2a\iff2\sin^2a=1-\cos2a$
