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?
$$2\cos x+3\cos y-2=0$$
$$2\sin x+3\sin y-8=0$$

The equations can be rewritten as
$$a\cos x+b\cos y=c$$
$$a\sin x+b\sin y = d$$
Squaring both gives us
$$a^2\cos^2x+b^2\cos^2 y+2ab\cos x\cos y=c^2$$
$$a^2\sin^2 x+b^2\sin^2 y+2ab\sin x\sin y=d^2$$
Adding these together gives
$$a^2+b^2+2ab\cos(x-y)=c^2+d^2$$
Which allows us to solve for $\cos(x-y)$ in terms of given constants. Multiplying the first two equations by $\cos y$ and $\sin y$ respectively and adding them gives us
$$a\cos(x-y)+b=c\cos y+d\sin y$$
which allows you to solve for $y$.
