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

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By Expert Faculty of Sri Chaitanya

a

n π \pm \frac{π}{3}, n \in Z

b

2 n π \pm \frac{π}{3}, n \in Z

c

n π + (- 1)^{n} \frac{π}{3}, n \in Z

d

none of these

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detailed solution

Correct option is B

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

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

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

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

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

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

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