<s>[INST] 'Strange' trigonometric roots of $x^5-4x^4+2x^3+5x^2-2x-1$ - could someone explain? This quintic equation has $5$ real roots:
$$x^5-4x^4+2x^3+5x^2-2x-1=0 \tag{1}$$
The roots are, from left to right:
$$x_1=\frac{\cos \frac{19}{22} \pi}{\cos \frac{1}{22} \pi}$$
$$x_2=\frac{\cos \frac{9}{22} \pi}{\cos \frac{19}{22} \pi}$$
$$x_3=\frac{\cos \frac{7}{22} \pi}{\cos \frac{5}{22} \pi}$$
$$x_4=\frac{\cos \frac{1}{22} \pi}{\cos \frac{7}{22} \pi}$$
$$x_5=\frac{\cos \frac{5}{22} \pi}{\cos \frac{9}{22} \pi}$$
I found these roots numerically, using ISC. The equation was found on Wikipedia in a different form (for $x-4/5$), no solutions were provided.


I can derive this equation for each root individually. But I don't see how do all five roots 'fit' together.
Why ${1,5,7,9,19}$? Why there is no $3/22, 13/22$ or $17/22$ in any of the arguments?
In any case, I would be grateful for the explanation for how these roots all fit together.

 [/INST] Using the equality $\cos(\pi \pm x)=-\cos(x)$ you get
$$x_1=-\frac{\cos \frac{3 }{22} \pi}{\cos \frac{1}{22} \pi} \\
x_2=-\frac{\cos \frac{9}{22} \pi}{\cos \frac{3}{22} \pi} \\
x_3=-\frac{\cos \frac{15}{22} \pi}{\cos \frac{5}{22} \pi} \\
x_4=-\frac{\cos \frac{21}{22} \pi}{\cos \frac{7}{22} \pi} \\
x_5=-\frac{\cos \frac{27}{22} \pi}{\cos \frac{9}{22} \pi}
$$
Note that the identity 
$$\cos(3x)= \cos(x) [2 \cos(2x)-1]$$
gives you a nicer form for the roots. With this form, after the proper substitution you should be able to reduce your polynomial to the minimal polynomial of $\cos(\frac{\pi}{11})$, which will explain where the roots are coming.
</s>