Solving $x^5+x^4-12x^3-21x^2+x+5=0$ $$x^5+x^4-12x^3-21x^2+x+5=0$$ 
I think it can be solved by trigonometric ways, but how?

Inspecting the discriminant and the small primes where the polynomial splits suggests that its splitting field is the class field of modulus $31$ corresponding to the subgroup $\{\pm 1, \pm 5, \pm 6\}$ of $(\Bbb Z/31\Bbb Z)/\{\pm 1\}$.
Which means that the splitting field is $\Bbb Q\left(2\left(\cos\frac{2\pi}{31}+\cos\frac{10\pi}{31}+\cos\frac{12\pi}{31}\right)\right)$.
Then I computed the minimal polynomial of this quantity (pure curiosity), and ended up on your polynomial exactly.
So the roots are $2\left(\cos\frac{2\pi}{31}+\cos\frac{10\pi}{31}+\cos\frac{12\pi}{31}\right)$, and its conjugates $$2\left(\cos\frac{4\pi}{31}+\cos\frac{20\pi}{31}+\cos\frac{24\pi}{31}\right) , \\
2\left(\cos\frac{8\pi}{31}+\cos\frac{22\pi}{31}+\cos\frac{14\pi}{31}\right), \\
2\left(\cos\frac{16\pi}{31}+\cos\frac{18\pi}{31}+\cos\frac{28\pi}{31}\right),\\
2\left(\cos\frac{30\pi}{31}+\cos\frac{26\pi}{31}+\cos\frac{6\pi}{31}\right).$$
I have no idea on how to guess this without the heavy machinery.
