I dropped a parenthesis in my derivation, and missed a cos(phi) cancellation, which caused me to omit the odd pi solutions;
phi = n*p where n is an integer
You can reduce all the trig to
(1+s)*(1-es)^e
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ozo's solution is a trivial solution because the sine of any multiple of 2*pi is zero, the cosine is 1, and the tangent of half the angle (180) also is zero. So the entire RHS reduces to 1. Thus any integer multiple of 2pi is a solution.
If you multiply out the denominator you'll get the left hand term as a power of the constant e equal to a right hand term which does not involve e in the same degree, that is the tangent expression of phi over 2 does not contain an e term. This means that phi must be given in terms of e and therefore in non-trigometrical terms as well.
I suspect thate there is no simple algebraic solution.
I dropped a parenthesis in my derivation, and missed a cos(phi) cancellation, which caused me to omit the odd pi solutions;
phi = n*p where n is an integer
You can reduce all the trig to
(1+s)*(1-es)^e - (1+es)^(e+1) = 0 [for s!=1]
where s=sin(phi)
Sorry did not get the chance to try this but it is time to close this now. Thanks for the effort.
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