De Moivre Theorem. Find the exact values of the solutions of the equations. By considering $z=cos\theta +i \sin \theta$ and using de Moivre's theorem, show that $$\sin5 \theta=\sin \theta (16 \sin^4 \theta-20 \sin^2 \theta +5)$$
Find the exact values of the solutions of the equation $16x^4-20x^2+5=0$ 
I've no problem with the first part. 
Then i found the x is equal to 0, $\sin36, \sin108$ and $\sin144$. But the given answer 0 is not included. Why?
 A: You have probably realized that $\sin\theta$ takes all values so you can set $x=\sin\theta$. Then you will have
$$\sin5\theta = x(16x^4-20x^2 + 5)$$
This means that it's easy to find the roots for $x(16x^4-20x^2+5)$ being $x=\sin\theta$ whenever $\sin5\theta=0$ (ie $\theta = n\pi/5$).
Now you have ten candidates per period, but due to symmetry only half of them are distinct: $\sin0, \pm\sin\pi/5, \pm\sin2\pi/5$ (I guess here is where you went wrong, not identifying all roots). Now you know that the polynomial has these five roots and it's the roots of it's factors (for $x(16x^4-20x^2+5)$ to be zero either $x$ is zero or $(16x^4-20x^2+5)$ is zero). The polynomial $x$ has root $0$ and $16x^4-20x^2+5$ has (at most) four roots which must be the remaining four. 
So the solutions are: $x=\pm\sin\pi/5$ and $x=\pm\sin2\pi/5$.
As pointed out you could also note that it can be solved using the same method as the quadratic:
$$16x^4-20x^2+5 = (4x^2-5/2)^2 - 5/4$$
which will give the roots
$$4x^2-5/2 = {\pm\sqrt5\over2}$$
$$2x = \pm\sqrt{{5\over2}\pm{\sqrt5\over2}}$$
$$x = \pm\sqrt{{5\over8}\pm{\sqrt5\over8}}$$
as a bonus you got a value for $\sin\pi/5$ and $\sin2\pi/5$:
$$\sin\pi/5 = \sqrt{{5\over8}-{\sqrt5\over8}}$$
$$\sin2\pi/5 = \sqrt{{5\over8}+{\sqrt5\over8}}$$
A: *

*You missed the solution $x=\sin(72°)=\sqrt{{5\over8}+{\sqrt5\over8}}$, 

*you didn't consider the factor $x$ in front of the polynomial.
There are $5$ solutions in $x$ in total, leading to $10$ distinct solutions in $\theta$ (direct and supplementary angles), not counting periodicity.

