sin((n+1)x)/sin(x) can be expressed as a polynomial in cos(x): sin((n+1)x)/sin(x) = U_n(cos(x)) where U_n is the nth Chebyshev polynomial of the second kind. See Wikipedia for more: https://en.wikipedia.org/wiki/Chebyshev_polynomials.

In any case, that tells you that in principle sum_{n=1}^N sin(nx) = sin(x) f(cos(x)) for some polynomial f. You can also write sin(x) = sqrt(1 - cos(x)^2) to express everything in terms of cos. Let g(t) = f(t)sqrt(1-t^2). Then you are trying to solve g(cos(x)) = 1 so the result is an inverse trig function applied to the solutions of g(t) = 1.

The derivation above has cos^{-1} instead of tan^{-1}, but since you can express tan(x) = sqrt{1-cos(x)^2}/cos(x), and cos(x) = 1/sqrt(1 + tan(x)^2) you can express solutions in terms of the inverse tangent.

Now the task is up to you--calculate the degrees of the polynomials involved!