<s>[INST] Find the exact value of $\cos\frac{2\pi}{5}$ by solving equation There's this concept in the topic of complex numbers which I don't really understand much (and was devastated upon realising it'll appear in the topic often) - trigonometry!
I'm super lost to be honest.
We are asked to express $\cos3\theta$ and $\cos2\theta$ in terms of $\cos\theta$. 
Then show that $\cos3\theta=\cos2\theta$ can be written as $4z^3-2z^2-3z+1=0$ where $z=\cos\theta$.
Lastly, by solving the equation above for $z$, we are to find out the value of $\cos\frac{2\pi}{5}$.
I attempted the question with just subbing in $z=\cos\theta$ into the equation and I got: $$4\cos^3\theta-3\cos\theta=\cos2\theta$$ and don't really know where to go next.
 [/INST] Here I use the formulas
$$\cos(x+y) = \cos x \cos y - \sin x \sin y$$
$$\sin(x+y) = \sin x \cos y + \cos x \sin y.$$
So $$\cos 2\theta = \cos(\theta+\theta) = \cos^2 \theta -\sin^2 \theta = 2\cos^2 \theta - 1.$$
Similarly
\begin{align}
\cos 3\theta = \cos (2\theta+\theta) & = \cos 2\theta\cos \theta-\sin 2\theta \sin \theta \\
& = 2\cos^3 \theta-\cos \theta-2\sin^2 \theta \cos \theta \\
& = 4\cos^3 \theta -3\cos \theta.
\end{align}
Now if $\cos 3\theta = \cos 2\theta$ you get
$$4\cos^3 \theta - 2\cos^2 \theta - 3\cos \theta + 1 = 0$$
which is what you claim putting $z = \cos \theta$. It's easy to see that $z = 1$ is a solution, so 
$$4z^3-2z^2-3z+1 = (z-1)(4z^2+2z-1) = \bigg(z-1\bigg)\bigg(z+\frac{1+\sqrt{5}}{4}\bigg)\bigg(z-\frac{-1+\sqrt{5}}{4}\bigg). $$
Now $\theta = \frac{2\pi}{5}$ satisfies $\cos 3\theta = \cos 2\theta$, and $\cos \frac{2\pi}{5}$ must be positive (because $\frac{2\pi}{5} < \frac{\pi}{2}$) and different from $1$ (because $\frac{2\pi}{5} \neq 0$). Hence
$$\cos \frac{2\pi}{5} = \frac{\sqrt{5}-1}{4}.$$
</s>