<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] The trick here is to find $2$ representations of the complex exponential $$e^{ix} = \cos(x)+i\sin(x).$$ 
With this in mind, we have 
$$e^{2ix} = \cos(2x)+i\sin(2x)$$ and 
$$e^{3ix} = \cos(3x)+i\sin(3x) $$ along with 
$$(e^{ix})^2 = (\cos(x)+i\sin(x))^2 = (\cos(x)^2 - \sin(x)^2) + i(2\sin(x)\cos(x))$$ and 
$$(e^{ix})^3 = (\cos(x)+i\sin(x))^3 = (\cos(x)^3 -3\cos(x)\sin(x)^2) + i(3\cos(x)^2\sin(x) - sin(x)^3).$$
Equating the real part of $e^{2ix}$ yields 
$$\cos(2x) = \cos(x)^2 - \sin(x)^2.$$
Using the Pythagorean identity, we see that $\sin(x)^2 = 1-\cos(x)^2$ and so we have
$$\cos(2x) = \cos(x)^2 - \sin(x)^2 = \cos(x)^2 - (1-\cos(x)^2) = 2\cos(x)^2-1.$$
Similarly, we may equate the real part of $e^{3ix}$ to get 
$$\cos(3x) = \cos(x)^3 -3\cos(x)\sin(x)^2 = \cos(x)^3 - 3\cos(x)(1-\cos(x)^2) = 4\cos(x)^3 - 3\cos(x).$$
So, if we set $$\cos(2z) = \cos(3z)$$
then we have 
$$2\cos(z)^2-1 = 4\cos(z)^3 - 3\cos(z)$$ which implies 
$$4\cos(z)^3 - 2\cos(z)^2 -3 \cos(z) +1 =0, \text{ as desired}.$$
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