Hello:

I am trying to solve this system:

A*cos(x) + B*sin(y) = C
D*sin(x) + E*cos(y) = F

I have tried several substitutions, but it will always lead to a 4th degree equation. I habe also used identities, for example sin(y) = sqrt(1-cos(y)^2), sin(x)=(exp(i*x)-exp(i*x))/2*i and cos(x)=(exp(i*x)+exp(i*x))/2, and so on...

For instance:

A*cos(x) + B*sin(y) = C => sin(y)^2 = ((C-A*cos(x))/B)^2
D*sin(x) + E*cos(y) = F => cos(y)^2 = ((F-D*sin(x))/E)^2 = 1- sin(y)^2

So ((F-D*sin(x))/E)=^2 = 1 - ((C-A*cos(x))/B)^2

When replacing sin(x) = sqrt(1-cos(x)^2 and expanding, I will get a 4th degree equation.

HOWEVER: I remember that in highschool I used to solve this without having to solve a 4th degree equation. I do know the solution for a 4th degree equation, but as you may know, it's a BIG formula.

The solution for the system is:

y1= arcsin((M*G+K)/(G^2+H^2))
y2= arcsin((M*G-K)/(G^2+H^2))

x1= arccos((F-E*y1)/d)
x2= arccos((F-E*y2)/d)

Where M=C^2+B^2+F^2-A^2, G=2*F*E, H=-2*B*C

I got it like 10 years ago! but now I can't remember the trick. I need to document a program I wrote for solving equations.

Thanks!

Oh, I was just trying hard to solve the same problem.
The link
http://www.geometrictools.com/Documentation/IntersectionOfEllipses.pdf
seemed to indicate that you cannot avoid a 4th order polynomial.

So what you've written down is the actual solution to the problem???
That would be amazing.

I've tried calculations with the discriminant of the 4th order polynomial but it doesn't quite work yet.

What's the variables "d" and "K" anyway?