How solve this trigo system ?

[AlexFekken]

(09-18-2017 08:56 AM)ggauny@live.fr Wrote:  cos x - cos y = 1/2
sin x * sin y = 3/8

Alternatively, from a different angle :-), set x = p+q and y = p-q so that we get

-2sin(p)sin(q) = 1/2
(sin(p)cos(q) + cos(p)sin(q)) * (sin(p)cos(q) - cos(p)sin(q)) = 3/8

and eliminating cos^2 for sin^2 after simplifying the latter equation we get

-sin(p)sin(q) = 1/4
sin(p)^2-sin(q)^2 = 3/8

This leads to a quadratic equation in sin(p)^2 and/or sin(q)^2, etc.

So you can solve this exactly without the ugliness of dealing with a quartic....