<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] $$\begin{align}
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\cos(3\theta) - \cos(2\theta) &=
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\mathcal R \left\{e^{3i\theta}\right\} - \left\{e^{2i\theta}\right\} \\ \\ &=
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\mathcal R \left\{\left(e^{i\theta}\right)^3 - \left(e^{i\theta}\right)^2\right\} \\ \\ &=
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\mathcal R \left\{\left(\cos(\theta) + \rm i \sin(\theta) \right)^3 - \left(\cos(\theta) + \rm i \sin(\theta) \right)^2\right\} \\ \\ &=
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\mathcal R \left\{\begin{array} {ll}
& \left(\cos(\theta)^3 + 3~\rm i \cos(\theta)^2\sin(\theta) -
  3~\rm \cos(\theta)\sin(\theta)^2 - \rm i \sin(\theta)^2 \right)
\\ - & \left(\cos(\theta)^2 + 2~\rm i \cos(\theta) \sin(\theta) - \sin(\theta)^2\right)
\end{array}\right\} 
\\ \\ &=
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\cos(\theta)^3 - 3\cos(\theta)\sin(\theta)^2 - \cos(\theta)^2 + \sin(\theta)^2 \\ \\ &=
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z^3 - 3z(1 - z^2) - z^2 + 1 - z^2 \\ \\ &=
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4z^3 - 2z^2 - 3z + 1
\end{align}$$
</s>