If `sin theta, 1, cos 2 theta` are in G.P. , then `theta` =

Options

A. `n pi + (-1)^n "" pi/2`
B. `n pi + (-1)^(n-1) pi/2`
C. `2n pi`
D. none

Answer

To solve the problem where \( \sin \theta, 1, \cos 2\theta \) are in geometric progression (G.P.), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Condition for G.P.**:
For three numbers \( A, B, C \) to be in G.P., the condition is:
\[ B^2 = A \cdot C \]
Here, \( A = \sin \theta \), \( B = 1 \), and \( C = \cos 2\theta \). Thus, we have:
\[ 1^2 = \sin \theta \cdot \cos 2\theta \]
This simplifies to:
\[ 1 = \sin \theta \cdot \cos 2\theta \]

2. **Using the Double Angle Identity**:
We know from trigonometric identities that:
\[ \cos 2\theta = 1 - 2\sin^2 \theta \]
Substituting this into the equation gives:
\[ 1 = \sin \theta (1 - 2\sin^2 \theta) \]

3. **Rearranging the Equation**:
Expanding the equation leads to:
\[ 1 = \sin \theta - 2\sin^3 \theta \]
Rearranging gives us:
\[ 2\sin^3 \theta - \sin \theta + 1 = 0 \]

4. **Factoring the Polynomial**:
We can factor the polynomial:
\[ 2\sin^3 \theta - \sin \theta + 1 = 0 \]
This can be rewritten as:
\[ 2\sin^3 \theta - 2\sin^2 \theta + \sin^2 \theta + 1 = 0 \]
Grouping terms gives:
\[ (2\sin^2 \theta - 2\sin \theta + 1) + \sin \theta = 0 \]

5. **Finding Roots**:
We can set:
\[ \sin \theta (2\sin^2 \theta - 2\sin \theta + 1) + 1 = 0 \]
This gives us two cases:
- \( \sin \theta = -1 \)
- \( 2\sin^2 \theta - 2\sin \theta + 1 = 0 \)

6. **Solving for \( \sin \theta = -1 \)**:
From \( \sin \theta = -1 \):
\[ \theta = \frac{3\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]

7. **Analyzing the Quadratic Equation**:
The quadratic \( 2\sin^2 \theta - 2\sin \theta + 1 = 0 \) has a discriminant:
\[ D = (-2)^2 - 4 \cdot 2 \cdot 1 = 4 - 8 = -4 \]
Since the discriminant is negative, there are no real solutions from this quadratic.

### Final Answer:
Thus, the only solution for \( \theta \) is:
\[ \theta = \frac{3\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]