Solving this system of trigonometric equations How would I go about solving:
$$
\begin{cases}
\sin(x) + \cos(y) = \frac{1}{2}\\
\frac{\cos(x)}{\sin(y)} = -\frac{\sqrt{3}}{2}
\end{cases}
$$
Where
$$
x \in \left(\frac{-\pi}{6},\frac{7\pi}{6}\right) \, , \, \, y \in \left(\frac{-2\pi}{3},\frac{2\pi}{3}\right)
$$
I tried taking inverse trigonometric functions but it got super messy so I assume there must be a simpler way to solve this. Can I for example shift $\sin(x)$ by $\pi/2$ or some constant to obtain a cosine function?
 A: Hint: squaring both
$$
\begin{cases}
\sin^2 x=0.25-\cos y +\cos^2 y\\
\\cos^2 x=0.75 \sin^2 y
\end{cases}
$$
Adding both equations yields: $0.25\cos^2 y-\cos y=0$. Can you take it from here?
A: Let's just square the second equation, we will check at the end if squaring increases any solution.
$\frac{\cos^2x}{\sin^2y} = \frac34 \implies \sin^2x = 1 - \frac34\sin^2y$
$\implies \sin x = \pm \sqrt{1 - \frac{3}{4}\sin^2y}$
Substitute this in equation 1.
$\pm \sqrt{1 - \frac{3}{4}\sin^2y} = \frac{1}{2} - \cos y$
Square both sides and change $\sin^2x$ to $\cos^2x$ to get:
$4\cos y = \cos^2y$
$\implies \cos y = 0\text{ or }4$
$\implies \cos y = 0 \implies y =(2n + 1)\frac{\pi}{2} $
Substitute this in equation 1, $\sin x = \frac{1}{2} \implies x = n\pi + (-1)^{n}\frac{\pi}{6}$
From the given range of y and x in question,
$$y \in \{-\frac\pi 2, \frac\pi 2\}$$
$$x \in \{\frac\pi 6, \frac{5\pi}{6}\}$$
But according to the second equation, we know that $\cos x$ and $\sin y$ can't be of the same signs, this will reduce some solutions.
So, the ordered pair $(x,y)$ satisfying the above two equations are
$$(\frac\pi 6, -\frac\pi 2)$$
$$(\frac{5\pi}{6}, \frac\pi 2)$$
