<s>[INST] 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?
$$2\cos x+3\cos y-2=0$$
$$2\sin x+3\sin y-8=0$$
 [/INST] 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,
$$\cos(y - \theta) = \frac{b^2 + R^2 - a^2}{2 b R} $$
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