Given that, terms $\frac{1}{6} \sin \theta, \cos \theta$ and $\tan \theta$ are in GP.
$\begin{aligned}
& \therefore \quad \cos ^2 \theta=\frac{1}{6} \sin \theta \times \tan \theta \\
& \Rightarrow \quad \cos ^2 \theta=\frac{\sin ^2 \theta}{6 \cos \theta} \\
& \Rightarrow \quad 6 \cos ^3 \theta-\sin ^2 \theta=0 \\
& \Rightarrow \quad 6 \cos ^3 \theta-1+\cos ^2 \theta=0
\end{aligned}$
$\begin{aligned}
& \Rightarrow \quad 6 \cos ^3 \theta+1+\cos ^2 \theta=0 \\
& \Rightarrow \quad 6 \cos ^3 \theta+\cos ^2 \theta-1=0 \\
& \Rightarrow \quad 6 \cos ^3 \theta+4 \cos ^2 \theta-3 \cos ^2 \theta+2 \cos \theta-2 \cos \theta-1=0 \\
& \Rightarrow \quad 6 \cos ^3 \theta-3 \cos ^2 \theta+4 \cos ^2 \theta-2 \cos \theta+2 \cos \theta-1=0 \\
& \Rightarrow \quad 3 \cos ^2 \theta(2 \cos \theta-1)+2 \cos \theta(2 \cos \theta-1)+1(2 \cos \theta-1) \\
& \quad=0 \\
& \Rightarrow \quad(2 \cos \theta-1)\left(3 \cos ^2 \theta+2 \cos \theta+1\right)=0
\end{aligned}$
For $3 \cos ^2 \theta+2 \cos \theta+1=0$ value of $\cos \theta$ will be imaginary so only $2 \cos \theta-1=0$ will be considered to find $\cos \theta$.
$\begin{aligned}
& \Rightarrow \quad 2 \cos \theta-1=0 \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \cos \theta=\cos \frac{\pi}{3} \\
& \therefore \quad \theta=2 \mathrm{n} \pi \pm \frac{\pi}{3}
\end{aligned}$