How to find the exact value of $ \cos(36^\circ) $? [duplicate]

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 $\dfrac{1+\sqrt{5}}{4}.$

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, $\cos{(\pi/5)} = (1+\sqrt{5})/4$.