I have the following trigonometric system of equations.
$ \begin{align*} && \sin(x)\cos(y) &= \frac{1+\sqrt{3}}{4} \\ && \cos(x)\sin(y) &= \frac{-1+\sqrt{3}}{4} \\ \end{align*} $
I am trying to find a quick way to solve this system for $x$ and $y$ with $0<x,y<\frac{\pi}{2}$. I have managed to find one set of solutions ($x=\frac{\pi}{4}$ and $y=\frac{\pi}{12}$) but I am unable to find the other pair which I can clearly see after graphing the system.
I was able to deduce, by making use of the addition formula for $sin$, that:
$\begin{align*}
&&\sin(x + y) = \frac{\sqrt{3}}{2} \implies x + y = \frac{\pi}{3}
\end{align*}$
Using this fact, I found my pair of solutions by forming a quadratic in $\tan(x)$ after playing around with addition formula for $\tan$.
I am certain there is an easier way to solve the problem that I am missing. Any tips for solving this system would be greatly appreciated, thanks.
Hint:
$$\begin{align*} \sin(x + y) &= \frac{\sqrt{3}}{2} &\implies x + y = \left\{\frac{\pi}{3},\frac{2\pi}{3}\right\}+2\pi n\\ \sin(x - y) &= \frac{1}{2} &\implies x - y = \left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}+2\pi n\\ \end{align*} $$
Can you take it from here?