roots allows trig keyword: return cubic roots in terms of cos

This was motivated by a question at http://stackoverflow.com/questions/24740517/python-sympy-trigonometric-solution-of-cubic-equation.  A special class of cubics has real solutions that can either be expressed in terms of expressions containing sqrt(-1) -- which cancels out at evaluation -- or in terms of the transcendental functions `cos` and `acos`. Using the `trig` keyword allows the latter instead of the former (which is the default).
if `trig` is not True then this is what is obtained:

```
{2/3 + 13/(9*(-1/2 + sqrt(3)*I/2)*(8/27 + sqrt(237)*I/9)**(1/3)) +
(-1/2 + sqrt(3)*I/2)*(8/27 + sqrt(237)*I/9)**(1/3): 1, 2/3 + (-1/2 -
sqrt(3)*I/2)*(8/27 + sqrt(237)*I/9)**(1/3) + 13/(9*(-1/2 -
sqrt(3)*I/2)*(8/27 + sqrt(237)*I/9)**(1/3)): 1, 2/3 + 13/(9*(8/27 +
sqrt(237)*I/9)**(1/3)) + (8/27 + sqrt(237)*I/9)**(1/3): 1}
```

I'll commit this in 12+h if there is no objection.