If `sintheta,1,cos2theta` are in G.P., then find the general values of `theta`

Explanation

To solve the problem where \( \sin \theta, 1, \cos 2\theta \) are in geometric progression (G.P.), we need to use the condition for three numbers to be in G.P. The condition states that if \( a, b, c \) are in G.P., then \( b^2 = ac \).

### Step 1: Set Up the Equation
Let:
- \( a = \sin \theta \)
- \( b = 1 \)
- \( c = \cos 2\theta \)

According to the G.P. condition:
\[ 1^2 = \sin \theta \cdot \cos 2\theta \]
This simplifies to:
\[ 1 = \sin \theta \cdot \cos 2\theta \]

### Step 2: Substitute for \( \cos 2\theta \)
Using the double angle identity for cosine, we have:
\[ \cos 2\theta = 1 - 2\sin^2 \theta \]
Substituting this into our equation gives:
\[ 1 = \sin \theta (1 - 2\sin^2 \theta) \]

### Step 3: Rearrange the Equation
Expanding the equation:
\[ 1 = \sin \theta - 2\sin^3 \theta \]
Rearranging it leads to:
\[ 2\sin^3 \theta - \sin \theta + 1 = 0 \]

### Step 4: Factor the Cubic Equation
We can factor the cubic equation:
\[ 2\sin^3 \theta - \sin \theta + 1 = 0 \]
This can be factored as:
\[ (\sin \theta + 1)(2\sin^2 \theta - 2\sin \theta + 1) = 0 \]

### Step 5: Solve Each Factor
1. From the first factor:
\[ \sin \theta + 1 = 0 \implies \sin \theta = -1 \]
The general solution for \( \sin \theta = -1 \) is:
\[ \theta = \frac{3\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]

2. For the second factor:
\[ 2\sin^2 \theta - 2\sin \theta + 1 = 0 \]
The discriminant of this quadratic is:
\[ (-2)^2 - 4 \cdot 2 \cdot 1 = 4 - 8 = -4 \]
Since the discriminant is negative, there are no real roots from this factor.

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