If ` sin alpha , 1 , cos 2 alpha ` are in GP, then find the general solution for ` alpha `

Answer

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

### Step 1: Understanding the condition for GP
If three terms \( a, b, c \) are in GP, then the relationship between them is given by:
\[ b^2 = ac \]
In our case, we have:
\[ 1^2 = \sin \alpha \cdot \cos 2\alpha \]
This simplifies to:
\[ 1 = \sin \alpha \cdot \cos 2\alpha \]

### Step 2: Using the double angle formula
We know that:
\[ \cos 2\alpha = 1 - 2\sin^2 \alpha \]
Substituting this into our equation gives:
\[ 1 = \sin \alpha (1 - 2\sin^2 \alpha) \]
Expanding this, we have:
\[ 1 = \sin \alpha - 2\sin^3 \alpha \]

### Step 3: Rearranging the equation
Rearranging the equation leads to:
\[ 2\sin^3 \alpha - \sin \alpha + 1 = 0 \]

### Step 4: Factoring the cubic equation
We can factor this equation. First, we can check for possible rational roots. By testing \( \sin \alpha = -1 \):
\[ 2(-1)^3 - (-1) + 1 = -2 + 1 + 1 = 0 \]
Thus, \( \sin \alpha = -1 \) is a root. We can factor out \( (1 + \sin \alpha) \):
\[ 2\sin^3 \alpha - \sin \alpha + 1 = (1 + \sin \alpha)(2\sin^2 \alpha - 2\sin \alpha + 1) \]

### Step 5: Analyzing the quadratic factor
Now we need to analyze the quadratic:
\[ 2\sin^2 \alpha - 2\sin \alpha + 1 \]
To find its discriminant:
\[ D = b^2 - 4ac = (-2)^2 - 4 \cdot 2 \cdot 1 = 4 - 8 = -4 \]
Since the discriminant is negative, this quadratic has no real roots. Therefore, the only solution comes from:
\[ 1 + \sin \alpha = 0 \implies \sin \alpha = -1 \]

### Step 6: Finding the general solution for \( \alpha \)
The sine function equals -1 at:
\[ \alpha = \frac{3\pi}{2} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]
This can also be expressed as:
\[ \alpha = 4n\pi - \frac{\pi}{2} \quad \text{for } n \in \mathbb{Z} \]

### Final Answer
The general solution for \( \alpha \) is:
\[ \alpha = 4n\pi - \frac{\pi}{2} \quad \text{where } n \in \mathbb{Z} \]