Given, $4\cos 2\theta \cos 3\theta =\sec \theta$

$\Rightarrow 4\cos 2\theta ·\cos 3\theta =\frac{1}{\cos \theta }$

$\Rightarrow 4\cos 2\theta \cos 3\theta \cos \theta =1$

$\Rightarrow 2\cos 2\theta \left(2\cos 3\theta \cos \theta \right)=1$

$\Rightarrow 2\cos 2\theta \left(\cos 4\theta +\cos 2\theta \right)=1$

$\Rightarrow 2\cos 2\theta \cos 4\theta +2{\cos }^{2}2\theta -1=0$

$\Rightarrow 2\cos 2\theta \cos 4\theta +\cos 4\theta =0$

$\Rightarrow \cos 4\theta \left(2\cos 2\theta +1\right)=0$

Then, $\cos 4\theta =0\Rightarrow \theta =\left(2n+1\right)\frac{\mathrm{\pi }}{8}, n\in I ...\left(\mathrm{i}\right)$

$\theta =\frac{\mathrm{\pi }}{8}, \frac{3\mathrm{\pi }}{8}, \frac{5\mathrm{\pi }}{8}, \frac{7\mathrm{\pi }}{8}$

And, $2\cos 2\theta +1=0\Rightarrow \cos 2\theta =-\frac{1}{2}$

$\theta \in \frac{\mathrm{n\pi }}{2}\pm \frac{\mathrm{\pi }}{6}, n\in I ...\left(\mathrm{ii}\right)$

$\theta =\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{3},\frac{2\mathrm{\pi }}{3}$

Hence, we have total $7$ solutions.