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? 

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.
