I forgot to mention this is over the interval [0,2pi). For 4cos(4x)sin(x) + sin(4x)cos(x) I can easily see that at 0 and pi the sin portions will produce solutions, but I can't think of a way to find where else
4cos(4x)sin(x) = -sin(4x)cos(x)
especially since the cos(4x) and cos(x) will never both equal 0 for the same angle. I have tried reducing cos(4x) and sin(4x) down using half angle formulas, but the end result just seems more messy. I was able to change both equations into forms of tan, with the first equation being:
-tan(4x) = 4tan(x)
This just feels like a step in the wrong direction though.