How to solve the equation $ x^{13} = 1$ by radicals? (And $x^{19}=1$) Is there any elementary way to solve the equation $ x^{13}= 1 $ by means of radicals? If not, how to get all the solutions?
Remark: The transcendental form of the solution by means of sines and cosines is not allowed, but only radicals, since this equation is solvable.
The solutions in terms of sines and cosines are $ \cos (k\pi/13)+i\sin(k\pi/13) $ for $ k = 0.1... 12$. So, what I say is that I am looking for an explicit  solution in terms of radicals. The trivial answer $\sqrt[13]{1} $ is also excluded.
 A: First, we establish some ground rules. All cubics and quartics can be solved by radicals, agreed? Thus, given,
$$\frac{x^{13}-1}{x-1}=x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=0\tag1$$
to solve for $x$ in radicals all we have to do is to "break-up" the $12$-deg into quadratics, cubics, or quartics. 
I. Method 1
Equivalently, $(1)$ is the palindromic quartic,
$$F(x)=x^4 + u x^3 + (u^2 - u - 1)x^2 + u x + 1 =0\tag2$$
with its coefficients determined by,
$$u^3 - u^2 - 4u - 1 = 0\tag3$$
$F(x)=0\,$ has $4$ roots, but as there are $3$ choices of $u$, using each one will yield the $4\times3=12$ roots of $(1)$.  As proof, eliminating $u$ between $(2),(3)$ will recover the $12$-deg. The WA command is, 

Resultant[x^4 + u x^3 + (u^2 - u - 1)x^2 + u x + 1, u^3 - u^2 - 4u - 1, u]

II. Method 2
For primes $p=6n+1$, we can "factor" the cyclotomic polynomial into $n$ palindromic sextics with coefficients determined by an equation of degree $n$. But a palindromic sextic is just a cubic in disguise since given,
$$x^6 + a x^5 + b x^4 + c x^3 + b x^2 + a x + 1=0$$
then one can solve it as $x+1/x=y$ where $y$ are the roots of the cubic,
$$y^3+a y^2 + (b-3) y +(-2a+c)= 0$$
$p=13:$

Resultant[1 - u x + 2 x^2 - (1 + u) x^3 + 2 x^4 - u x^5 + x^6, -3 + u + u^2, u]

$p=19:$

Resultant[1 - u x - (2 - u^2) x^2 - (6 + 2 u - u^2) x^3 - (2 - u^2) x^4 - u x^5 + x^6, -7 - 6 u + u^2 + u^3, u]

and so on. Thus, you only need to solve $-3 + u + u^2=0$ and $-7 - 6 u + u^2 + u^3=0$, respectively. (Click link for more on $p=19$). For $p=31$, you'll have to tackle the quintic $-5 + u + 21 u^2 - 12 u^3 - u^4 + u^5=0$. But that's another story.
A: $x^{13}=1$ implies $x=\sqrt[13]{1}=\zeta_{13}^n$ for any integer $n$ where $\zeta_{13}$ is a primitive 13th root of unity. 
The 13 distinct numbers $\zeta_{13}^n$ are the vertices of a regular 13-gon with one vertex at 1. Now you can find the internal angle of a regular 13-gon and apply trigonometry. Not sure if this is what your looking for, but I hope it helps. 
A: Why sine and cosine are not allowed? $sin(\pi/2)$ for example is algebraic.
So the solutions are
$$
x_k=\omega^k, \quad \omega=\cos(2\pi/13)+i\sin(2\pi/13)
$$
A: Here’s an unsatisfactory contribution to the discussion:
The Galois group of the extension $\Bbb Q(\zeta_{13})\supset\Bbb Q$ is cyclic of order $12$, and so has cyclic quotients of order three and four. So there are intermediate fields $K_3$ and $K_4$ of degree three and four over $\Bbb Q$, both cyclic. Our field does not contain the cube roots of unity, nor the fourth roots. But once you adjoin these, you can now express $K_3(\zeta_3)$ as gotten from $\Bbb Q(\zeta_3)$ by adjunction of a cube root, and you can express $K_4(i)$ as gotten from $\Bbb Q(i)$ by adjunction of a fourth root.
