We have,
$$
\begin{aligned}
& \Rightarrow \cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{5 \pi}{15} \cos \frac{7 \pi}{15} \cos \frac{30 \pi}{15}=x \\
& \Rightarrow \cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{\pi}{3} \cos \frac{7 \pi}{15} \cos 2 \pi=x \\
& \Rightarrow \quad \cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{7 \pi}{15}=2 x \\
& {\left[\because \cos \frac{\pi}{3}=\frac{1}{2} \text { and } \cos 2 \pi=1\right]}
\end{aligned}
$$


$$
\begin{aligned}
& \Rightarrow \cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \left(\pi-\frac{8 \pi}{15}\right)=2 x \\
& \Rightarrow \quad-\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15}=2 x \\
& \Rightarrow-\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos 2^2 \frac{\pi}{15} \cdot \cos 2^3 \cdot \frac{\pi}{15}=2 x \\
& \Rightarrow \quad-\frac{\sin 2^4 \frac{\pi}{15}}{24} \sin \frac{\pi}{15} \\
& \Rightarrow \quad\left[\because \cos A \cos 2 A \cos 2^2 A \ldots \cos 2^{14-1} A=\frac{\sin 2^{14} A}{2^n \sin A}\right] \\
& \Rightarrow 2 x=-\frac{\sin \frac{16 \pi}{15}}{16 \sin \frac{\pi}{15}} \Rightarrow 2 x=-\frac{\sin \left(\pi+\frac{\pi}{15}\right)}{16 \sin \frac{\pi}{15}}=\frac{\sin \frac{\pi}{15}}{16 \sin \frac{\pi}{15}} \\
& \Rightarrow 2 x=\frac{1}{16} \Rightarrow x=\frac{1}{32} \Rightarrow \frac{1}{x}=32 \Rightarrow \frac{1}{8 x}=4
\end{aligned}
$$