It is well known that sin(30deg)=1/2.  For brevity let z=sin(6); then sin(6deg) is a root of 1/2 = 5z - 20z^3 + 16z^5. 

By the small angle approximation sin(x)~=x (in radians) then sin(6deg) = sin(pi/30) ~= pi/30 = 0.105, so whichever root of the polynomial is the root, it is the one close to 0.105.

Rearranging the polynomial a bit gives (32z^5-1) - (20z^3-5z) = 0.  From this it is easy to factor out one term: (2z-1) * (16z^4+8z^3-16z^2-8z+1) = 0  But 2z-1=0 implies z=1/2 which is not the root we seek.

Then I plugged the quartic into WolframAlpha, and saw the roots were expressible in terms of quadratic radicals.  So I knew that the quartic 16x^4+8x^3-16x^2-8x+1=0 can be expressed in the form of a squared quadratic equals a squared linear.

So a bit of reverse engineering from the radical forms of the roots, I concluded the important intermediate form is:

(4z^2+z-3/2)^2 = 5*(z+1/2)^2

Then solving the two resulting quadratic equations we get

z= (1/8)*(-1-sqrt[5]-sqrt[30-6sqrt[5]]) = -0.91355
z= (1/8)*(-1-sqrt[5]+sqrt[30-6sqrt[5]]) = 0.10453
z= (1/8)*(-1+sqrt[5]-sqrt[30+6sqrt[5]]) = -0.66913
z= (1/8)*(-1+sqrt[5]+sqrt[30+6sqrt[5]]) = 0.97815

The second one fits our expected approximation, so sin(6deg) = (1/8)*(-1-sqrt[5]+sqrt[30-6sqrt[5]]) = 0.10453.

--- addendum: Solving the quartic ---

Solving 16x^4+8x^3-16x^2-8x+1=0. 

"Most" of the quartic will equal a squared quadratic.

Expand (Ax^2 + Bx + C)^2 into A^2x^4 + 2ABX^3 + other stuff.  Then A=4 and B=1 are quickly found

We want the difference between (4x^2 + x + C)^2 and 16x^4+8x^3-16x^2-8x+1 to be the square of a linear.  Then

(4x^2 + x + C)^2 - (16x^4+8x^3-16x^2-8x+1) = (8C+17)x^2 + (2C+8)x + (C^2-1).

Now the right side needs to be the square of a linear, but that happens exactly when the discriminant equals zero.  Then

(2C+8)^2 - 4*(8C+17)*(C^2-1) = 0.

This simplifies to 8C^3+16C^2-16C-33.  This polynomial has a factorization of (2C-3)*(4C^2+14C+11), one root is rational: C=3/2.

Then finally we can substitute C=3/2 and get that our starting quartic of 16x^4+8x^3-16x^2-8x+1=0 can be rearranged into (4z^2+z-3/2)^2 = 5*(z+1/2)^2.