Lots of cubic equations can be solved in terms of trigonometric roots, and they do not have to involve angles divisibie by $\pi/7$ or $\pi/9$.
The minimal denominator greater than $9$ for such an equation is $13$. For simplicity I shall render $2\cos(2m\pi)$$2\cos(2m\pi/13)$ as $[m]$ below.
We start with the cyclic-sum relation
$[0]+[1]+[2]+...+[12]=0.$
As $[0]=2$ and $[13-m]=[m]$ this reduces to
$[1]+[2]+[3]+[4]+[5]+[6]=-1,$
and we group these terms into three pairs having the same ratio $\bmod13$:
$([1]+[5])+([2]+[3])+([4]+[6])=-1$
where $5/1\equiv 2/3\equiv 4/6\equiv 5\bmod 13$.
Then $x_1=[1]+[5],x_2=[2]+[3],x_3=[4]+[6]$ have the sum $-1$, and using this sum combined with the trigonometric sum-product ratios we find that the other symmetric sum-product combinations are also integers:
$([1]+[5])([2]+[3])=[1][2]+[1][3]+[5][2]+[5][3]$
$=[1]+[3]+[2]+[4]+[3]+[7]+[2]+[8]$
$=[1]+2[2]+2[3]+[4]+[5]+[6]$
$([1]+[5])([4]+[6])=2[1]+[2]+[3]+[4]+2[5]+[6]$
$([2]+[3])([4]+[6])=[1]+[2]+[3]+2[4]+[5]+2[6]$
$\therefore ([1]+[5])([2]+[3])+([1]+[5])([4]+[6])+([2]+[3])([4]+[6])=4([1]+[2]+[3]+[4]+[5]+[5]+[6])=-4$
and by similar manipulations
$([1]+[5])([2]+[3])([4]+[6])=2[0]+5([1]+[2]+[3]+[4]+[5]+[6])=-1$
So
$x_1=2[\cos(2\pi/13)+\cos(10\pi/13)]$
$x_2=2[\cos(4\pi/13)+\cos(6\pi/13)]$
$x_3=2[\cos(8\pi/13)+\cos(12\pi/13)]$
solve the equation
$x^3+x^2-4x+1=0.$