I'm not sure why you would consider that "hard". You can solve the second equation for x as a quadratic function of y. Putting that into the first equation gives a cubic equation for y.
Indeed. Do you have any strategies to find a root? There are many you can try. Once you find one, you can divide and just solve a quadratic.
Here is what I have done so far: But these solutions contains "cos". It is possible to rewrite the equations to the formula and then find solutions without any trigonometrical components. That is my problem...
Would have been a simpler computation for the program to have changed the x into 1/y rather than the other way around, small numbers to deal with overall. Were you expected to solve this cubic analytically and it's roots actually looked like that?
Yes. It does not matter how you start. You will get the same solutions with the same trigonometrical components. Plz help me find solutions without any trigonometrical components.
Is it just because the answer drops our nicely if that is the case, or do you somehow know it as a fact :S ?