Solve $\sin(5A) + \cos(5A)\sin(A) - \cos(3A) = 0$ How do you solve this equation for A:
$~~\sin(5A) + \cos(5A)\sin(A) - \cos(3A) = 0$
I've tried expanding it many times, but I can't seem to be able to reduce it to a format I can work with. Is there a simpler method of solution than repeated expansion?

$$ \sin(5A) + \cos(5A) \sin(A) - \sin(3A) = 0 $$
Let $ x = e^{iA} $ and use De Moivre's,
$$ \frac{x^5 - x^{-5}}{2i} + \frac{x^5 + x^{-5}}{2} \frac{x - x^{-1}}{2i} - \frac{x^3-x^{-3}}{2i} = 0 $$
Multiply by $ 4i x^6 $,
$$ 2(x^{11} - x) + (x^{10} + 1)(x^2 - 1) - 2(x^9 - x^3) =  $$
$$ (x^2 - 1)(x^{10} + 2x^9 + 2x + 1) = 0 $$
The phase of each root to the polynomial above (the ones with $ | x | = 1 $ at least) is a solution $ A $ to your equation (up to an integer addition of $ 2\pi $).
