Find coordinates of a point Q on the graph $\sin (x) + \cos (y) = 0.5$ given that the gradient of its tangent is perpendicular to point P. Note:
Point $P$ is on the $y$-axis and above the $x$-axis
$\frac{-\pi}{6}\le x \le\frac{7\pi}{6}$
$\frac{-2\pi}{3}\le y\le\frac{2\pi}{3}$
What I have done so far:
Solving for $P$: 
$$x = 0
\\ \sin (0) + \cos (y) = 0.5
\\ 0 + \cos (y) = 0.5
\\ y= \pm\frac{\pi}{3} $$
For $P$, $y \gt 0$
$\therefore y = \frac{\pi}{3}$
Solving for $\frac{dy}{dx}$: 
$$\sin(x) + \cos(y) = 0.5
\\ \cos(x) - \sin(y)\frac{dy}{dx} = 0$$
$\therefore \frac{dy}{dx} = \frac{\cos(x)}{\sin (y)}$
Derivative at $P$:
$$\frac{dy}{dx} = \frac{\cos(x)}{\sin (y)}
= \frac{\cos(0)}{\sin(\frac{\pi}{3})}
= \frac{2}{\sqrt3}$$
As for the gradient of the tangent line at $Q$ is perpendicular to that at $P$:
$\frac{dy}{dx} = \frac{-\sqrt3}{2}$
How do I solve for the coordinates of $Q$ after this?

Good work. Note the point $Q$ lies on the curve as well as the tangent line. So:
$$\begin{cases}\frac{\cos x}{\sin y}=-\frac{\sqrt{3}}{2}\\ \sin x+\cos y=0.5 \end{cases} \Rightarrow$$
From the first:
$$\cos^2x=\frac34(1-\cos^2y)\Rightarrow \cos y=\pm\sqrt{1-\frac43\cos^2x}$$
Now sub it to the second:
$$1-\frac43\cos^2x=\frac14-\sin x+\sin^2x\Rightarrow \\
4\sin^2x+12\sin x-7=0 \Rightarrow \sin x=\frac12.$$
Referring to the given constraints, the final answer is:
$$x=\pi-\frac{\pi}{6},y=\frac{\pi}{2}$$
