Finding the roots, with multiplicity, of a polynomial with compound angle formula My question is to find the roots, counted with multiplicity, of the polynomial equation 
$16x^5-20x^3+5x-1=0$ using the compound angle formula $\sin\left(5\theta\right)=16\sin^5\theta-20\sin^3\theta+5\sin\theta$
So after substituting $x=\sin\theta$, I get to the equation
$\sin(5\theta)=1$
Then I get an infinitude of $\theta$ values, which when I find the sines of these values, all correspond to the distinct solutions $x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(-\frac{3\pi}{10}\right)$.
What I don't understand is how I can then find which roots are repeating, since the degree of the polynomial is 5 hence there must be 5 roots when counted with multiplicity.
Also if possible, is there a way to solve this polynomial using the given compound angle formula without the need to find the distinct roots and then determine the ones which repeat?
This is because the solutions to this question say $x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{9\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{17\pi}{10}\right)$ without any reasoning, which makes me suspect I am unaware of some related theorem.

As already explained, the solutions are:
$$x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{9\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{17\pi}{10}\right)$$ 
Because,
$$\sin(5\theta)=1 \Rightarrow \theta=\frac{1}{5}\left(\frac{\pi}{2}+2k\pi\right)$$
and setting $k=0,1,2,3,4$ we get all five solutions. 
But $\sin\left(\frac{\pi}{10}\right)=\sin\left(\frac{9\pi}{10}\right)$ and $\sin\left(\frac{13\pi}{10}\right)=\sin\left(\frac{17\pi}{10}\right)$.
So the five roots (with multiplicity) are
$$1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right)$$.
Let's explain better why if make any other choice of $k$ we will always find an angle congruent to one of those in $S=\{\pi/10,\pi/2,9\pi/10,13\pi/10,17\pi/10\}$.
We can write $k=5n+r$ with $r \in \{0,1,2,3,4\}$ and $n \in \Bbb Z$. Replacing that at the expression of $\theta$ we get:
$$\theta=\frac{\pi}{10}+\frac{2k\pi}{5}=\frac{\pi}{10}+\frac{2(5n+r)\pi}{5}=\frac{\pi}{10}+2n\pi+\frac{2r\pi}{5} \equiv \frac{\pi}{10}+\frac{2r\pi}{5} $$
The last equivalence means that the angle $\frac{\pi}{10}+2n\pi$ is congruent to $\frac{\pi}{10}$. That means that both angle stop at the same point on the trigonometric circle and then if we apply any trigonometric function at those angles we will get the same value. The algebric value are different but geometricaly they represent the same point at the circle.
It means that if we want to find the different solution for $\sin\theta$ is enough to take  
$$\frac{\pi}{10}+\frac{2r\pi}{5}$$
with $r=0,1,2,3,4$.
