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 :)

Let $t=\frac{\pi}{5}$ (so $t$ is $36^\circ$). Since $108=180-72$, we have $3t=\pi-2t$ and therefore 
$$\sin(3t)=\sin(\pi-2t).$$
But $\sin(\pi-2t)=\sin(2t)=2\sin t\cos t$.
Also, by the identity you were given, $\sin(3t)=3\sin t-4\sin^3 t$. Thus
$$3\sin t-4\sin^3 t=2\sin t\cos t.$$
But $\sin t\ne 0$, so we can cancel a $\sin t$ and obtain
$$3-4\sin^2 t=2\cos t.$$
Substitute $1-\cos^2 t$ for $\sin^2 t$ and simplify a bit. We get
$$4\cos^2 t-2\cos t-1=0.$$
Use the Quadratic Formula to solve this quadratic equation for $\cos t$, rejecting the negative root. We get
$$\cos t=\frac{2+\sqrt{20}}{8}.$$
We can simplify this to $\dfrac{1+\sqrt{5}}{4}.$
