Find all $Z\in C$ such that $|z|=2$ and $Im(z^6) = 8Im(z^3).$ Practice problem from a college textbook.
Find all $Z\in\mathbb C$ such that $|z|=2$ and $Im(z^6) =  8Im(z^3).$
My method is:
$2^6\sin6\phi = 8\times2^3\sin3\phi$
$\sin6\phi = \sin3\phi$
$\sin3\phi = 2\sin3\phi\cos3\phi$
$\cos3\phi = \frac{1}{2}$
$3\phi = \frac{\pi}{3} , \frac{5\pi}{3}$
Yielding:
$\phi_{1,2,3} = \frac{\frac{\pi}{3}+2k\pi}{3} , k=0,1,2$
$\phi_{4,5,6} = \frac{\frac{5\pi}{3}+2k\pi}{3} , k=0,1,2$
This turned out wrong, because the solution provided by the textbook is
$z_1 = 2cis\frac{7}{9}\pi , z_2 = 2cis\frac{5}{9}\pi ,z_3 = 2cis\frac{2}{3}\pi .$
There are only three solutions according to the textbook.
Where did I go wrong?

The key equation here, as you write, is $$\sin 6\phi-\sin3\phi=0.$$
We can apply the formula for the difference of $\sin$ functions:
$$\sin 6\phi-\sin3\phi=2\sin(3\phi/2)\cos(9\phi/2)=0,$$
which immediately gives us all possible solutions:
$$\sin(3\phi/2)=0\Rightarrow \phi = \frac 23 \pi k,\quad k\in \Bbb Z, $$
$$\cos(9\phi/2)=0\Rightarrow \phi = \frac 19 \pi  +\frac 29 \pi k,\quad k\in \Bbb Z, $$
so, by limiting $\phi$ to the interval $[0,2\pi)$ we obtain possible values
$$0,\quad\frac 23 \pi ,\quad\frac 43 \pi,$$
and
$$\frac 19 \pi,\quad   \frac 13 \pi,\quad\frac 59 \pi,\quad\frac 79 \pi,\quad \pi,\quad\frac {11}{9} \pi,\quad\frac {13}{9} \pi,\quad\frac {5}{3} \pi,\quad\frac {17}{9} \pi.$$
You can also write these solutions as 
$$0,\quad\pm\frac 23 \pi,\quad\pm\frac 19 \pi,\quad   \pm\frac 13 \pi,\quad\pm\frac 59 \pi,\quad\pm\frac 79 \pi,\quad \pi.$$
