If 16sinx,cosx,tanx are in G.P then x is equal to

16sinx.tanx=cos2x⇒6cos2x=sin2xcosx⇒6cos3x=1−cos2x⇒6cos3x+cos2x−1=0⇒cosx−12(6cos2x+4cosx+2)=0⇒cosx=12x=2nπ±π3.n∈Z

If 16sinx,cosx,tanx are in G.P then x is equal to

16sinx.tanx=cos2x⇒6cos2x=sin2xcosx⇒6cos3x=1−cos2x⇒6cos3x+cos2x−1=0⇒cosx−12(6cos2x+4cosx+2)=0⇒cosx=12x=2nπ±π3.n∈Z

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

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$n π \pm \frac{π}{3}, n \in Z$

$n π + {(- 1)}^{n} \frac{π}{3}, n \in Z$

$2 n π \pm \frac{π}{6}, n \in Z$

$2 n π \pm \frac{π}{3}, n \in Z$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{1}{6} \sin x . \tan x = \cos^{2} x \\ \Rightarrow 6 \cos^{2} x = \frac{\sin^{2} x}{\cos x} \\ \Rightarrow 6 \cos^{3} x = 1 - \cos^{2} x \\ \Rightarrow 6 \cos^{3} x + \cos^{2} x - 1 = 0 \end{array}$

$\Rightarrow (\cos x - \frac{1}{2}) (6 \cos^{2} x + 4 \cos x + 2) = 0$

$\Rightarrow \cos x = \frac{1}{2}$

$x = 2 n π \pm \frac{π}{3} . n \in Z$

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