If `1/6 sin theta, cos theta` and `tan theta` are in G.P. then the general solution for `theta` is

Answer

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

### Step 1: Understand the condition for G.P.
For three terms \( a, b, c \) to be in G.P., the condition is:
\[ b^2 = ac \]
In our case, \( a = \frac{1}{6} \sin \theta \), \( b = \cos \theta \), and \( c = \tan \theta \).

### Step 2: Set up the equation
Using the G.P. condition:
\[ \cos^2 \theta = \left(\frac{1}{6} \sin \theta\right) \tan \theta \]
Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we can substitute this into the equation:
\[ \cos^2 \theta = \left(\frac{1}{6} \sin \theta\right) \left(\frac{\sin \theta}{\cos \theta}\right) \]
This simplifies to:
\[ \cos^2 \theta = \frac{1}{6} \frac{\sin^2 \theta}{\cos \theta} \]

### Step 3: Multiply both sides by \( 6 \cos \theta \)
To eliminate the fraction, we multiply both sides by \( 6 \cos \theta \):
\[ 6 \cos^3 \theta = \sin^2 \theta \]

### Step 4: Use the Pythagorean identity
We know that \( \sin^2 \theta = 1 - \cos^2 \theta \). Substituting this into the equation gives:
\[ 6 \cos^3 \theta = 1 - \cos^2 \theta \]

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

### Step 6: Factor the polynomial
We can factor this cubic equation. Let's assume it can be factored as:
\[ (2 \cos \theta - 1)(3 \cos^2 \theta + 2 \cos \theta + 1) = 0 \]

### Step 7: Solve for \( \cos \theta \)
From \( 2 \cos \theta - 1 = 0 \):
\[ \cos \theta = \frac{1}{2} \]
The other factor \( 3 \cos^2 \theta + 2 \cos \theta + 1 = 0 \) has no real solutions (as its discriminant is negative).

### Step 8: Find general solutions for \( \theta \)
The solution \( \cos \theta = \frac{1}{2} \) gives:
\[ \theta = \frac{\pi}{3} + 2n\pi \quad \text{and} \quad \theta = -\frac{\pi}{3} + 2n\pi \]
where \( n \) is any integer.

### Final General Solution
Thus, the general solution for \( \theta \) is:
\[ \theta = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z} \]

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