We have,
$$
\begin{aligned}
& 4 \cos 2 \theta \cdot \cos 3 \theta=\sec \theta \text {, (where } 0 < \theta < \pi) \\
& \Rightarrow \quad 2(2 \cos 2 \theta \cdot \cos 3 \theta)=\frac{1}{\cos \theta} \\
& \Rightarrow 2[\cos 5 \theta+\cos (-\theta)]=\frac{1}{\cos \theta} \\
& \quad \quad[\because 2 \cos A \cos B=\cos (A+B)+\cos (A-B)] \\
& \Rightarrow \quad 2[\cos 5 \theta+\cos \theta]=\frac{1}{\cos \theta} \quad[\because \cos (-\theta)=\cos \theta] \\
& \Rightarrow \quad 2 \cos 5 \theta \cdot \cos \theta+2 \cos ^2 \theta=1 \\
& \Rightarrow \quad(\cos 6 \theta+\cos 4 \theta)+\left(2 \cos ^2 \theta-1\right)=0 \\
& \Rightarrow \quad \cos 6 \theta+\cos 4 \theta+\cos 2 \theta=0 \\
& \Rightarrow \quad 2 \cos 4 \theta \cdot \cos 2 \theta+\cos 4 \theta=0 \\
& \Rightarrow \quad \cos 4 \theta(2 \cos 2 \theta+1)=0 \\
& \Rightarrow \quad \cos 4 \theta=0 \text { and } 2 \cos 2 \theta+1=0 \\
& \Rightarrow \quad 4 \theta=(2 n+1) \frac{\pi}{2}, \cos 2 \theta=-\frac{1}{2} \\
& \Rightarrow \quad \theta=(2 n+1) \frac{\pi}{8}, 2 \theta=2 \pi / 3 \text { or } \frac{4 \pi}{3} \\
& \Rightarrow \quad \theta=\frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}, \frac{\pi}{3}, \frac{2 \pi}{3}
\end{aligned}
$$
So, number of solution of given equation is 6 .