`"If cos" 2 theta, 1 "and sin" theta "are in GP., then "theta =`

Options

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

Explanation

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

### Step 1: Understanding the GP Condition
In a geometric progression, the middle term squared is equal to the product of the other two terms. Therefore, we can write the condition as:
\[ (\cos 2\theta)^2 = 1 \cdot \sin \theta \]

### Step 2: Simplifying the Equation
Since \( 1 \) is simply \( 1 \), we can rewrite the equation as:
\[ \cos^2 2\theta = \sin \theta \]

### Step 3: Using the Cosine Double Angle Identity
Using the identity \( \cos 2\theta = 1 - 2\sin^2 \theta \), we can substitute for \( \cos^2 2\theta \):
\[ \cos^2 2\theta = (1 - 2\sin^2 \theta)^2 \]

### Step 4: Expanding the Equation
Now, we expand the left-hand side:
\[ (1 - 2\sin^2 \theta)^2 = 1 - 4\sin^2 \theta + 4\sin^4 \theta \]
Thus, we have:
\[ 1 - 4\sin^2 \theta + 4\sin^4 \theta = \sin \theta \]

### Step 5: Rearranging the Equation
Rearranging gives us:
\[ 4\sin^4 \theta - 4\sin^2 \theta - \sin \theta + 1 = 0 \]

### Step 6: Substituting \( y = \sin \theta \)
Let \( y = \sin \theta \). The equation becomes:
\[ 4y^4 - 4y^2 - y + 1 = 0 \]

### Step 7: Finding Roots of the Polynomial
We can use synthetic division or factorization to find the roots of this polynomial. Testing \( y = -1 \):
\[ 4(-1)^4 - 4(-1)^2 - (-1) + 1 = 4 - 4 + 1 + 1 = 2 \quad \text{(not a root)} \]
Testing \( y = 1 \):
\[ 4(1)^4 - 4(1)^2 - (1) + 1 = 4 - 4 - 1 + 1 = 0 \quad \text{(a root)} \]

### Step 8: Factoring the Polynomial
Using \( y - 1 \) as a factor, we can divide the polynomial:
\[ 4y^4 - 4y^2 - y + 1 = (y - 1)(4y^3 + 4y^2 + 3y + 1) \]

### Step 9: Solving the Remaining Polynomial
We can solve \( 4y^3 + 4y^2 + 3y + 1 = 0 \) using numerical methods or further factorization.

### Step 10: Finding \( \theta \)
For the root \( y = 1 \):
\[ \sin \theta = 1 \implies \theta = \frac{\pi}{2} + 2n\pi, \quad n \in \mathbb{Z} \]

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

---