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?

This is an old question, but should be interesting to answer. If you want an aesthetic version, then the solution to,
$$x^5 + x^4 - 12x^3 - 21x^2 + x + 5 = 0$$
is given by,
$$x = \frac{1}{5}\left(-1+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\right)$$
and the $z_i$ are the roots of the quartic,
$$z^4 + 12679 z^3 + 78678031 z^2 + 362989005529 z + 31^{10} = 0$$
Notice how the quartic has a constant term that is a fifth power. This factors over the extension $\sqrt{5}$. So alternatively,
$$x = \frac{1}{5}\left(\beta_0-1+\frac{31}{\beta_0}+\frac{31}{\alpha}+\alpha\right)$$
where,
$$\alpha = \left(\tfrac{31}{4}\right)^{1/5}\left(-409-125\sqrt{5}+5\sqrt{-10(925-409\sqrt{5})}\right)^{1/5}\tag1$$
$$\beta_0 = \left(\tfrac{31}{4}\right)^{1/5}\left(-409+125\sqrt{5}+5\sqrt{-10(925+409\sqrt{5})}\right)^{1/5}\tag2$$

Added:

Israel's Maple answer can be simplified. The five roots $x_k$ for $k=0,1,2,3,4$ are,
$$x_k = \frac{1}{5}\left(\frac{1}{\beta_k^{-1}}-\frac{1}{\beta_k^0}+\frac{31}{\beta_k^1}+\frac{31a}{\beta_k^2}+\frac{31b}{\beta_k^3} \right)$$
where,
$$a=\frac{\beta_0^2}{\alpha}=\tfrac{1}{4}\left(11+5\sqrt{5}+\sqrt{-10(25-11\sqrt{5})}\right)$$
$$b=\frac{\alpha\,\beta_0^3}{31}=\tfrac{1}{4}\left(-1-5\sqrt{5}+\sqrt{-10(1525-\sqrt{5})}\right)$$
$$\beta_k = e^{2\pi\,i\,k/5}\,\left(31ab\right)^{1/5}\tag3$$
So it turns out that the root $(2)$ can be factored as $(3)$. Maple's answer also has the advantage that only one fifth root extraction of a complex number is needed.
