How to find the exact value of $ \cos(36^\circ) $? The problem reads as follows:

Noting that $t=\frac{\pi}{5}$ satisfies $3t=\pi-2t$, find the exact value of 
  $$\cos(36^\circ)$$

it says that you may find useful the following identities:
$$\cos^2 t+\sin^2 t = 1,\\
\sin 2t = 2\sin t\cos t,\\
\sin 3t = 3\sin t - 4\sin^3 t.
$$
Do I have to do a system of linear equations in function of ..what? $t$? $\cos$?
Thanks in advance :)

Here is a way using roots of unity.
We have $\cos 36 = \frac{\omega + \omega^{-1}}{2}$ where $\omega = \text{exp} \left ( \frac{2 i\pi}{10} \right )$.
We have $-w$ is a primitive $5^{th}$ root of unity, since $ -\exp(\frac{2i\pi}{10}) = \exp(i\pi)\exp(\frac{2i\pi}{10}) = \exp(\frac{12i\pi}{10}) = \exp(\frac{2ki\pi}{5})$, so it follows $\omega^4 - \omega^3 + \omega^2 - \omega + 1 = 0$.
Now, $x^4 - x^3 + x^2 - x + 1 = x^2(x^2 - x + 1 - \frac{1}{x} + \frac{1}{x^2}) = x^2\left ( \left (x + \frac{1}{x} \right )^2 - \left (x + \frac{1}{x} \right ) -1 \right )$.
Thus $0 = \omega^2 \cdot ((2 \cos 36)^2 - (2 \cos 36) - 1)$.
Therefore $4 \cos^3 36 - 2 \cos 36 - 1 = 0$.
Using the quadratic formula we then arrive at $$\cos 36 = \frac{1 + \sqrt{5}}{4}$$
