<s>[INST] Use $\cos 5\theta$ to find the roots of $x(16x^4 - 20x^2 + 5) = 0$ I used $\cos(3\theta + 2\theta)$ to prove the first part, but I don't know how to the $2$nd part.
Show that $\cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta,$ and hence show that
$$\text{the roots of }x(16x^4 - 20x^2 + 5) \text{ are: } 0,\cos\frac{\pi}{10}, \cos\frac{3\pi}{10},\cos\frac{7\pi}{10}, \cos\frac{9\pi}{10}$$
 [/INST] *

*The equation $$\;x(16x^4 - 20x^2 + 5) = 0\; \tag 1$$ can have $5$ real solutions (incl. multiplicity).

*From $\;\cos 5\theta = 16\cos^5 \theta  - 20\cos^3 \theta + 5\;$ we deduce that we will look for solutions of $(1)$ in the form of $\cos \theta\;$ where $\;\cos 5\theta = 0.$ 

*$\cos 5\theta = 0$ holds for $10$ different values of $\;\theta \in (-\pi,\pi]\;$ obtained when solving $$\cos 5\theta =\frac \pi2 +2k\pi\quad \text{or} \quad \cos 5\theta = -\frac \pi2 + 2k\pi,\; k=0,1,\dots,4.$$ 
These are $$\frac{\pi}{10},\; \frac{5\pi}{10},\; \frac{9\pi}{10},\;\frac{13\pi}{10},\;\frac{18\pi}{10}\;\text{and}\;-\frac{\pi}{10},\;\frac{3\pi}{10},\;\frac{7\pi}{10},\;\frac{11\pi}{10},\;\frac{15\pi}{10},\;$$
between them two and two have equal cosines (can you check which are the pairs?).
The solutions are those listed in the question.
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