<s>[INST] 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? 
 [/INST] 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 
$$
\begin{array}{c}
16t^4+8t^3-12t^2-4t+1=0\\
(2t+1)(8t^3-6t+1)=0\\
\end{array}
$$
</s>