Solution of $4 \cos x(\cos 2x+\cos 3x)+1=0$ Find the solution of the equation:
$$4 \cos x(\cos 2x+\cos 3x)+1=0$$
Applying trigonometric identity leads to 
$$\cos (x) \cos \bigg(\frac{x}{2} \bigg) \cos \bigg(\frac{5x}{2} \bigg)=-\frac{1}{8}$$
But I can't understand what to do from here. Could some suggest how to proceed from here? 

Hint:
$1)$ Note that:

*

*$\cos{2x}=2\cos^2{x}-1$, and

*$\cos{3x}=4\cos^3{x}-3\cos{x}$
$2)$ Using this, setup a cubic equation in $\cos{x}$.
$3)$ Now, solve the quartic equation by making the substitution $t=\cos{x}$.
$4)$ Finally, back-substitute and use $\cos{x}=\cos{\alpha} \Rightarrow x=2n\pi \ \pm \alpha$, where $n \in \mathbb{Z}$

Edit(Showed working of hint)
Using the identities, we get
$16\cos^4x+8\cos^3x-12\cos^2x-4\cos x+1=0$
Set $t=\cos x$.
Our equation is
$16t^4+8t^3-12t^2-4t+1=0 \Rightarrow (2t+1)(8t^3-6t+1)=0$
So, $2t+1=0$ or $8t^3-6t+1=0$
Back-substituting we get,

*

*$2\cos{x}+1=0 \Rightarrow \cos{x}=-1/2=\cos(2\pi/3)$
$\therefore x=2n\pi \pm 2\pi/3$, where $n \in \mathbb{Z}$


*$8\cos^3{x}-6\cos{x}+1=0$
Using the identity $\cos{3x}=4\cos^3{x}-3\cos{x}$, we get $2\cos{3x}+1=0 \Rightarrow \cos{3x}=-1/2=\cos(2\pi/3)$
$ \therefore 3x=2n\pi \pm 2\pi/3 \Rightarrow x=2\pi/3 \pm 2\pi/9$, where $n\in \mathbb{Z}$
Thus, $x=2\pi/3 \pm 2\pi/3$ or $x=2\pi/3 \pm 2\pi/9$
