The question is
Solve the equation $z^3=-8$
My attempt
I attempt to write it out in polar co-ordinates
since $z = r(\sin(x) + i\sin(x)) \\ z^3 = r^3(\cos(3x) + i\sin(3x))$
then $r^3\sin(3x) = -8$ and $r^3\cos(3x) = 0$
but from here I'm not really sure where to go , I've searched up the solution to this before and people have written $r^3 = 8$ so $\cos(3x) = -1$ and $\sin(3x) = 0$ .
By the fundamental theorem of algebra, we expect there to be 3 solutions.
If $z$ is a solution to $z^3 = -8$, then $|z^3| = |-8| = 8$, but since $|ab| = |a| * |b|$ and $|z|$ is a real number $\geq 0$, this implies $|z|^3 =8$ so $|z| = 2$.
Therefore we have the magnitude of all solutions to $z^3 = -8$. Now, we just need to match the angle. We have $(r e^{i \theta})^3 = r^3 e^{3 i \theta}$, and we already know that $r = 2$, so we just need $3 \theta = \pi (\mod 2 \pi)$.
$\theta = \pi/3$ is obviously a solution, for the other two solutions, add $2 \pi/3$ and $4 \pi/3$.
Therefore, the solutions are $2 e^{i \pi/3}, 2 e^{3 \pi i / 3} = -2$ and $2 e^{5 \pi i/3}$.
You are on the right track to write $$ z^3=r^3(\cos(3\theta)+i\sin(3\theta)),\quad (r\ge 0)\;. $$ But you should observe that $z^3=-8$ implies by comparing the real and imaginary parts that $$ r^3\cos(3\theta) = -8,\quad r^3\sin(3\theta)=0\;.\tag{1} $$
So you have $$ r^3=8,\quad \cos(3\theta)=-1,\quad \sin(3\theta)=0\;. $$ Consequently, you get $r=2$. To find out $\theta$, note that $$ \cos(3\theta)=\cos(\pi) $$ which gives you $$ 3\theta=-\pi+2k\pi\quad\text{ and thus }\quad \theta =-\frac{\pi}{3}+\frac{2k\pi}{3} $$ for some integer $k$. Now each pair of $(r,\theta)$ gives you a solution $$ z=r\cos(3\theta)\;. $$ By periodicity of the consine function, you really have three solutions.
Alternatively, you may also solve the equation by factoring: $$ z^3+8=(z+2)(z^2-2z+4) $$
$$z^3=-8$$ $$z^3 = r^3\cdot[\cos(3\cdot\theta) + i\sin(3\cdot\theta)]$$
Let's substitute $R = r^3$, $\rho= 3\cdot\theta$ so that:
$$R\cdot\sin(\rho) = 0$$ $$R\cdot\cos(\rho) = -8$$
Since $|z^3| = R = |-8| = 8$ you can get $\rho$ by doing: $$8\cdot\cos(\rho) = -8$$ $$\cos(\rho) = -1$$ $$\rho = (2k-1)\cdot\pi$$ where $k$ is an arbitrary integer $k \in \mathbb{Z}$. Let's make the anti-substitution and we obtain:
$$r = \sqrt[3]{R}=2$$ $$\theta = \frac{2k-1}{3}\cdot\pi$$ So, $$z = 2\cdot\left[\cos\left(\frac{2k-1}{3}\cdot\pi\right) + i\sin\left(\frac{2k-1}{3}\cdot\pi\right)\right]$$
Although it seems that there are infinite solutions for $z(k)$, this is not true, because $\cos()$ and $\sin()$ are periodic. Because of the fundamental theorem of algebra $z^3+8=0$ will have 3 solutions, which also matches with the $\cos()$ and $\sin()$ periodicity for $z$. Thus, possible solutions are defined this way:
For $k=0$, $z = 2\cdot\left[\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right] = 1-\sqrt{3}\cdot i$.
For $k=1$, $z = 2\cdot\left[\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right] = 1+\sqrt{3}\cdot i$.
For $k=2$, $z = 2\cdot[\cos(\pi) + i\sin(\pi)] = -2$
We could also apply the "triple-angle" identities to write $ \ \sin(3 \theta) \ = \ \sin \theta · (3 - 4 \sin^2 \theta) \ = \ 0 \ \ $ and $ \ \cos(3 \theta) \ = \ 4 \cos^3 \theta - 3 \cos \theta \ = \ -1 \ \ . $ From the first of these, we have either $ \ \sin \theta \ = \ 0 \ \Rightarrow \ \theta \ = \ 0 \ , \ \pi \ \ $ or $ \ \sin^2 \theta \ = \ \frac34 \ \Rightarrow \ \cos^2 \theta \ = \ \frac14 \ \ . $ Inserting these results into the second equation, we find that $ \ \theta \ = \ 0 \ $ is not a solution, but $ \ \theta \ = \ \pi \ $ is $ \ ( \ \cos(3 \pi) \ = \ 4 · [-1]^3 - 3 ·[-1] \ = \ -1 \ ) \ . $ For $ \ \cos \theta \ = \ \pm \frac12 \ \ , $ we see that $ \ 4 · \left[-\frac12 \right]^3 - 3 ·\left[-\frac12 \right] \ = \ +1 \ \ , $ but $ \ 4 · \left[+\frac12 \right]^3 - 3 ·\left[+\frac12 \right] \ = \ -1 \ \ . $
Hence the three angle solutions are $ \ \theta \ = \ \pi \ , \ \pm \ \frac{\pi}{3} \ \ , $ which gives the three solutions to the original equation (the three complex cube-roots of $ \ -8 \ $ as $$ 2·(cis \ \pi) \ \ = \ \ -2 \ \ \ , \ \ \ 2·\left(cis \ \pm \frac{\pi}{3} \right) \ \ = \ \ 1 \ \pm \ i\sqrt3 \ \ . $$