Solving equation of type $a\cos x+b\cos y-c=0$ and $a\sin x+b\sin y-d=0$ Here's the questions
There are two equations:
$a\cos x+b\cos y-c=0 $ and $ a\sin x+b\sin y-d=0$ .
For instance 
What is the value of $x$ and $y$ in following question?
$$2\cos x+3\cos y-2=0$$
$$2\sin x+3\sin y-8=0$$

just a hint
From the first equation
$$2\cos (x)=2-3\cos (y) $$
and from the second
$$2\sin (x)=2-3\sin (y) $$
thus
$$(2\sin (x))^2+(2\cos (x))^2=$$
$$4=8-12\cos(y )-12\sin(y)+9$$
$$\cos(y)+\sin (y)=\frac {13}{12}$$
$$\sqrt {2}\cos (y-\frac {\pi}{4})=\frac {13}{12}$$
You can finish.
