If `1/6 sinx, cosx, tan x` are in G.P. then x=,

Options

A. `n pi +- (pi)/(3), n in Z`
B. `2n pi +- (pi)/(3), n in Z`
C. `n pi + (-1)^(n) (pi)/(3), n in Z`
D. none of these

Answer

To solve the problem where \( \frac{1}{6} \sin x, \cos x, \tan x \) are in geometric progression (G.P.), we can follow these steps:

### Step 1: Set up the G.P. condition
For three terms \( a, b, c \) to be in G.P., the condition is:
\[ b^2 = ac \]
In our case, we have:
\[ b = \cos x, \quad a = \frac{1}{6} \sin x, \quad c = \tan x \]
Thus, we can write:
\[ \cos^2 x = \left(\frac{1}{6} \sin x\right) \tan x \]

### Step 2: Substitute \( \tan x \)
Recall that \( \tan x = \frac{\sin x}{\cos x} \). Substituting this into our equation gives:
\[ \cos^2 x = \left(\frac{1}{6} \sin x\right) \left(\frac{\sin x}{\cos x}\right) \]
This simplifies to:
\[ \cos^2 x = \frac{1}{6} \frac{\sin^2 x}{\cos x} \]

### Step 3: Multiply through by \( \cos x \)
To eliminate the fraction, multiply both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)):
\[ \cos^3 x = \frac{1}{6} \sin^2 x \]

### Step 4: Use the Pythagorean identity
We know from the Pythagorean identity that \( \sin^2 x = 1 - \cos^2 x \). Substitute this into the equation:
\[ \cos^3 x = \frac{1}{6} (1 - \cos^2 x) \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 6 \cos^3 x + \cos^2 x - 1 = 0 \]

### Step 6: Let \( y = \cos x \)
Let \( y = \cos x \). The equation becomes:
\[ 6y^3 + y^2 - 1 = 0 \]

### Step 7: Solve the cubic equation
To solve \( 6y^3 + y^2 - 1 = 0 \), we can use numerical methods or factorization techniques. By trial and error, we can check for rational roots. Testing \( y = \frac{1}{2} \):
\[ 6\left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^2 - 1 = 6 \cdot \frac{1}{8} + \frac{1}{4} - 1 = \frac{3}{4} + \frac{1}{4} - 1 = 0 \]
Thus, \( y = \frac{1}{2} \) is a root.

### Step 8: Find the general solution for \( x \)
Since \( y = \cos x = \frac{1}{2} \), we find:
\[ x = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \]
The general solution for \( x \) is:
\[ x = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z} \]

### Final Answer
Thus, the values of \( x \) are:
\[ x = 2n\pi + \frac{\pi}{3} \quad \text{or} \quad x = 2n\pi - \frac{\pi}{3}, \quad n \in \mathbb{Z} \]