If $\frac{1}{6}\sin x,\cos x,\tan x$  are in $G.P$., then$r$ is equal to

Solution:

It is given that $\frac{1}{6}\sin x,\cos x,\tan x$ are in $GP$. 

$\therefore {\cos }^{2}x=\frac{1}{6}\sin x t\mathrm{an}x$

$\Rightarrow 6{\cos }^{2}x=\sin x \tan x$

$\begin{array}{l}\Rightarrow 6{\cos }^{3}x+{\cos }^{2}x-1=0\\ \Rightarrow \left(\cos x-\frac{1}{2}\right)\left(6{\cos }^{2}x+4\cos x+2\right)=0\end{array}$

$\Rightarrow \cos x=\frac{1}{2}\left[\because {\cos }^{2}x+4\cos x+2=0\text{ has imaginary roots }\right]$

$\begin{array}{l}\Rightarrow \cos x=\cos \frac{\pi }{3}\\ \Rightarrow x=2n\pi \pm \frac{\pi }{3},n\Rightarrow Z.\end{array}$