If  $\frac{1}{6}\sin \theta ,\cos \theta$ and  $\tan \theta$  are in GP, and range of $\theta$ is $\left[0,n\pi \right]$ and  has 5 values, then n =

Since, $\frac{1}{6}\sin \theta ,\cos \theta ,\tan \theta$ are in G.P.

$\therefore {\cos }^2\theta =\frac{1}{6}\sin \theta .\tan \theta$

${\cos }^2\theta =\frac{1}{6}\frac{{\sin }^2\theta }{\cos \theta }$  put ${\sin }^2\theta =1-{\cos }^2\theta$

$\Rightarrow 6{\cos }^3\theta +{\cos }^2\theta -1=0$

Note that  $\cos \theta =\frac{1}{2}$ satisfies the equation (by trial)

$\therefore (2\cos \theta -1)(3{\cos }^2\theta +2\cos \theta +1)=0$

$\Rightarrow \cos \theta =\frac{1}{2}$           (other $\cos \theta$ values of  are imaginary)

$\Rightarrow \cos \theta =\cos \frac{\pi }{3}$

$\Rightarrow \theta =2n\pi \pm \frac{\pi }{3},n\in Z$

$\Rightarrow for n=0, \theta =\pm \frac{\mathrm{\pi }}{3}$

$$\begin{gathered} \begin{array}{cl} & \\ \therefore \mathrm{\theta }=\frac{\mathrm{\pi }}{3},2\mathrm{\pi }-\frac{\mathrm{\pi }}{3},2\mathrm{\pi }+\frac{\mathrm{\pi }}{3},4\mathrm{\pi }-\frac{\mathrm{\pi }}{3} \mathrm{and} 4\mathrm{\pi }+\frac{\mathrm{\pi }}{3}\mathrm{are} \mathrm{the} 5 \mathrm{values} \mathrm{taken} \\ \end{array} \end{gathered}$$

$$\begin{gathered} \mathrm{by} \mathrm{\theta } \mathrm{in} \left[0,5\mathrm{\pi }\right] \\ \Rightarrow \mathrm{n}=5 \end{gathered}$$

$for n=1, \theta =2\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3} \mathrm{for} \mathrm{n}=2, \mathrm{\theta }=4\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3}$