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? 
 A: 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 $x=2\pi/3 \pm 2\pi/9$
A: Using the identities 
$$
\cos(2x)=2\cos^2x-1\qquad \cos(3x)=4\cos^3x-3\cos x
$$
yields
$$
4\cos x(2\cos^2x-1+4\cos^3x-3\cos x)+1=0
$$
that is 
$$
16\cos^4x+8\cos^3x-12\cos^2x-4\cos x+1=0
$$
Set $t=\cos x$. Our equation is 
$$
\begin{array}{c}
16t^4+8t^3-12t^2-4t+1=0\\
(2t+1)(8t^3-6t+1)=0\\
\end{array}
$$
A: Using  Werner Formula,
$$-1=2(\cos x+\cos3x+\cos2x+\cos4x)$$
Using $\sum \cos$ when angles are in arithmetic progression  as $\sin\dfrac x2\ne0,$ as $\sin\dfrac x2=0\implies x=2n\pi$ where $n$ is any integer
$$-\sin\dfrac x2=\sin\dfrac{9x}2-\sin\dfrac x2\iff\sin\dfrac{9x}2=0$$
$$\implies\dfrac{9x}2=m\pi$$ where $m$ is any integer
But if $9\mid m, \sin\dfrac x2=0,$ so we need $9\nmid m$
A: Thinking about the answer, we might notice that if $\theta=\frac{2\pi k}9$, then $\cos9\theta=1$. We can write this as
$$\begin{align}\cos9\theta-1&=4(4\cos^3\theta-3\cos\theta)^3-3(4\cos^3\theta-3\cos\theta)-1\\
&=(16\cos^4\theta+8\cos^3\theta-12\cos^2\theta-4\cos\theta+1)^2(\cos\theta-1)=0\end{align}$$
From this we can see that the solutions were all the solutions to $\cos9\theta=1$ except for $\cos\theta=1$. If $\theta=\frac{\pm2\pi}3$, then $\cos\theta=-\frac12$, and if $\theta=\frac{2\pi(3k\pm1)}9$, then $\cos3\theta=-\frac12$ so all the cases in the big factor are taken into account by $(2\cos\theta+1)(2\cos3\theta+1)=0$.  
Indeed we can go back to the original equation and find that $4\cos\theta\cos2\theta=2(\cos(1+2)\theta+\cos(1-2)\theta)=2\cos3\theta+2\cos\theta$, so it reads
$$\begin{align}4\cos x(\cos2x+\cos3x)+1&=2\cos3x+2\cos x+4\cos3x\cos x+1\\
&=(2\cos3x+1)(2\cos x+1)=0\end{align}$$
Thus if we could have seen through this at the outset, we could have made quick work of the problem.
