<s>[INST] 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 :)
 [/INST] To find $\cos{\pi/5}$, note that
$$\sin{(3 \pi/5)} = \sin{(2 \pi/5)}$$
and
$$\sin{(3 \pi/5)} = 3 \sin{(\pi/5)} - 4 \sin^3{(\pi/5)} = 2 \sin{(\pi/5)} \cos{(\pi/5)}$$
Thus
$$2 \cos{(\pi/5)} = 3 - 4 \sin^2{(\pi/5)} = 4 \cos^2{(\pi/5)} - 1$$
Let $y=\cos{(\pi/5)}$.  Then
$$4 y^2-2 y-1=0 \implies y = \frac{1 + \sqrt{5}}{4}$$
because $y>0$.  Thus, $\cos{(\pi/5)} = (1+\sqrt{5})/4$.
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