Solving equation of type $a\cos x+b\cos y-c=0$ and $a\sin x+b\sin y-d=0$

Question

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

Answer 1

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.

Answer 2

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, $$\cos(y - \theta) = \frac{b^2 + R^2 - a^2}{2 b R} $$

Answer 3

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 $y$.