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

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(A) Given that $\frac{1}{6} \sin 	heta, \cos 	heta, 	an 	heta$ are in $G.P.$Therefore,$(\cos 	heta)^2 = (\frac{1}{6} \sin 	heta) 	imes (	an

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

Vedclass · Answer

(A) Given that $\frac{1}{6} \sin 	heta, \cos 	heta, 	an 	heta$ are in $G.P.$Therefore,$(\cos 	heta)^2 = (\frac{1}{6} \sin 	heta) 	imes (	an 	heta)$$\cos^2 	heta = \frac{1}{6} \sin 	heta 	imes \frac{\sin 	heta}{\cos 	heta}$$\cos^3 	heta = \frac{1}{6} \sin^2 	heta$$\cos^3 	heta = \frac{1}{6} (1 - \cos^2 	heta)$$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: $6(\frac{1}{8}) + \frac{1}{4} - 1 = \frac{3}{4} + \frac{1}{4} - 1 = 0$.So,$\cos 	heta = \frac{1}{2} = \cos(\frac{\pi}{3})$.The general solution is $	heta = 2n\pi \pm \frac{\pi}{3}, n \in Z$.

If $\frac{1}{6} \sin \theta, \cos \theta, \tan \theta$ are in $G.P.$,then the general solution of $\theta$ is

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