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?
$$2\cos x+3\cos y-2=0$$
$$2\sin x+3\sin y-8=0$$
 A: 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.
A: I'll assume $a > 0$ and $b > 0$.
Let $R = \sqrt{c^2 + d^2}$, so that $c = R \cos(\theta)$ and $d = R \sin(\theta)$ for some $\theta \in [0,2\pi)$.  The points $[0,0]$, $[a \cos(x), a \sin(x)]$ and $[c, d]$ form a triangle with sides of length $a$, $b$ and $R$.
We need $R \le a + b$, $a \le R + b$ and $b \le R + a$ for this to be possible.
By the law of cosines we get 
$$\cos(x - \theta) = \frac{a^2 + R^2 - b^2}{2 a R} $$
Similarly,
$$\cos(y - \theta) = \frac{b^2 + R^2 - a^2}{2 b R} $$
A: 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 $y$.
