If frac{1}{6} sin theta, cos theta and tan th | vedclass

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(B) Given that $\frac{1}{6} \sin 	heta, \cos 	heta, 	an 	heta$ are in $GP$. Therefore,$\cos^2 	heta = \frac{1}{6} \sin 	heta \cdot 	an 	heta$.

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$2 n \pi \pm \frac{\pi}{3}$

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(B) Given that $\frac{1}{6} \sin 	heta, \cos 	heta, 	an 	heta$ are in $GP$. Therefore,$\cos^2 	heta = \frac{1}{6} \sin 	heta \cdot 	an 	heta$. $\cos^2 	heta = \frac{\sin^2 	heta}{6 \cos 	heta} \implies 6 \cos^3 	heta = \sin^2 	heta$. Since $\sin^2 	heta = 1 - \cos^2 	heta$,we have $6 \cos^3 	heta = 1 - \cos^2 	heta$. $6 \cos^3 	heta + \cos^2 	heta - 1 = 0$. Let $x = \cos 	heta$. Then $6x^3 + x^2 - 1 = 0$. By inspection,$x = \frac{1}{2}$ is a root since $6(\frac{1}{8}) + \frac{1}{4} - 1 = \frac{3}{4} + \frac{1}{4} - 1 = 0$. Dividing $6x^3 + x^2 - 1$ by $(2x - 1)$,we get $3x^2 + 2x + 1 = 0$. The discriminant of $3x^2 + 2x + 1$ is $D = 2^2 - 4(3)(1) = 4 - 12 = -8 Thus,$\cos 	heta = \frac{1}{2} = \cos \frac{\pi}{3}$. The general solution is $	heta = 2n\pi \pm \frac{\pi}{3}$.

If $\frac{1}{6} \sin \theta, \cos \theta$ and $\tan \theta$ are in geometric progression,then the solution set of $\theta$ is

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