Some exact values of $\cos \theta$ using de Moivre's theorem Presently I am faced with the following question:

By showing that $$\cos5\theta = \cos\theta(16\cos^4\theta - 20\cos^2\theta + 5)$$ and then solving the equation $\cos5\theta = 0$, deduce that 
$$\cos^2\left(\frac{\pi}{10}\right) = \frac{5+\sqrt{5}}{8}$$
and hence find the exact values of $\cos^2\left(\frac{3\pi}{10}\right), \cos^2\left(\frac{7\pi}{10}\right), \cos^2\left(\frac{9\pi}{10}\right)$.

I have proved the first part using de Movire's theorem and after solving $\cos5\theta = 0$ reached the following equation:
$$\cos^2 \left(\frac{\pi}{10}\right) = \frac{5\pm\sqrt{5}}{8}$$
Where $\frac{\pi}{10}$ can be replaced with  $\frac{3\pi}{10}$,  $\frac{7\pi}{10}$, etc. as they are other possible solutions when finding $\theta$ from $\cos5\theta = 0$.
However I am unsure how I am supposed to know whether or not to take the positive or negative root dependent on the angle (i.e. for $\frac{\pi}{10}$, $\frac{7\pi}{10}$,  $\frac{3\pi}{10}$,  $\frac{9\pi}{10}$), in order to get the exact value.

Since $\cos^2 \theta$ is strictly decreasing on $\left[0, \frac{\pi}{2}\right]$, we must have that
$$\cos^2 \left(\frac{\pi}{10}\right) > \cos^2 \left(\frac{3 \pi}{10}\right),$$ and so
$$\cos^2 \left(\frac{\pi}{10}\right) = \frac{5 + \sqrt{5}}{8} \qquad \text{and} \qquad \cos^2 \left(\frac{3\pi}{10}\right) = \frac{5 - \sqrt{5}}{8}.$$
The remaining values can be determined by a similar argument, or by using these values together with the easy identity $\cos^2 \theta = \cos^2 (\pi - \theta)$.
