To solve a Trig equation To solve for $\cos{x}$ of equation $2\cos{3x}-4\cos{2x}+10\cos{x}-3=0.\quad$ Now using usual identities i reduced this to a 

$8\cos^3{x }-8\cos^2{x}+4\cos{x}+1=0$.

How do i proceed?

Thanks
Hint:

It can be shown the equation $p(t)=8t^3-8t^2+4t+1=0$ has only one real root between $-1$ and $0$, because


$p'(t)=24t^2-16t+44=4(6t^2-4t+1)$ has no real root, hence is positive for all $t$.
$p(-1)=-23$, $p(0)=1$, so we can apply the intermediate value theorem.


The rational root theorem yields no root (the candidates would be, taking into account the previous results, $-\frac12,-\frac14,-\frac18$).  We only can say the root is between $-\frac14$ and $-\frac18$.
 Thus the only exact method is Cardano's method:


First  eliminate the term in $t^2$, setting $z=2t-\frac23$, you should get the equation:
$$z^3+\frac23z+\frac{47}{27}=0 \qquad\text{(if I'm not mistaken)}.$$
Then set $z=u+v$ and add the condition $uv=-\frac29$ to simplify the resulting equation.
This leads to the (non-linear) system:
\begin{cases}u^3+v^3=-\frac{47}{27}\\u^3v^3=-\frac 8{243}\end{cases}
a classic problem on quadratic equations: $u^3$ and $v^3$ are the roots of 
$$U^2+\frac{47}{27}U-\frac{8}{243}=0.$$
HINT

You can set for example $n=\cos x$, so then $\cos^2 x$ would change to $n^2$ and so on...

Your equation will look like $8n^3-8n^2+4n+1=0
$

After this, solve like you would do a cubic equation considering different methods like factoring, etc..

At the end don't forget to put back $\cos x$ in the place of $n$ !!