Let S be the set of all `alpha in R` such that the equation, `cos 2x + alpha sin x = 2alpha -7` has a solution. The S, is equal to

Options

A. R
B. `[1,4]`
C. `[3,7]`
D. `[2,6]`

Explanation

To solve the equation \( \cos 2x + \alpha \sin x = 2\alpha - 7 \) and find the set \( S \) of all \( \alpha \in \mathbb{R} \) such that the equation has a solution, we can follow these steps:

### Step 1: Rewrite the equation using the double angle formula
We know that \( \cos 2x = 1 - 2\sin^2 x \). We can substitute this into the equation:

\[ 1 - 2\sin^2 x + \alpha \sin x = 2\alpha - 7 \]

### Step 2: Rearrange the equation
Rearranging the equation gives:

\[ -2\sin^2 x + \alpha \sin x + 1 = 2\alpha - 7 \]

This simplifies to:

\[ -2\sin^2 x + \alpha \sin x + 8 - 2\alpha = 0 \]

### Step 3: Form a quadratic equation
Let \( y = \sin x \). The equation can be rewritten as:

\[ -2y^2 + \alpha y + (8 - 2\alpha) = 0 \]

### Step 4: Use the discriminant
For the quadratic equation \( -2y^2 + \alpha y + (8 - 2\alpha) = 0 \) to have real solutions for \( y \), the discriminant must be non-negative. The discriminant \( D \) is given by:

\[ D = b^2 - 4ac = \alpha^2 - 4(-2)(8 - 2\alpha) \]

Calculating this gives:

\[ D = \alpha^2 + 8(8 - 2\alpha) = \alpha^2 + 64 - 16\alpha \]

### Step 5: Set the discriminant \( D \geq 0 \)
We need:

\[ \alpha^2 - 16\alpha + 64 \geq 0 \]

### Step 6: Factor the quadratic expression
The expression can be factored as:

\[ (\alpha - 8)^2 \geq 0 \]

### Step 7: Determine the values of \( \alpha \)
Since \( (\alpha - 8)^2 \geq 0 \) is always true for all \( \alpha \), the equality holds when \( \alpha = 8 \). Therefore, the set \( S \) includes all real numbers \( \alpha \):

\[ S = \mathbb{R} \]

### Conclusion
The set \( S \) of all \( \alpha \in \mathbb{R} \) such that the equation has a solution is:

\[ S = \mathbb{R} \]