For what values of a does the equation `cos^(2) 2x + a sin x = 2a - 7` poses a real solution ?

Options

A. `a lt 2`
B. `a ge 8`
C. `a gt 8`
D. a is integer `lt - 2`

Answer

To solve the equation \( \cos^2(2x) + a \sin x = 2a - 7 \) for the values of \( a \) that allow for real solutions, we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \cos^2(2x) + a \sin x = 2a - 7 \]
Using the identity \( \cos^2(2x) = 1 - 2\sin^2(x) \), we can rewrite the equation:
\[ 1 - 2\sin^2(x) + a \sin x = 2a - 7 \]

### Step 2: Rearrange the equation
Rearranging gives us:
\[ -2\sin^2(x) + a \sin x + 1 - 2a + 7 = 0 \]
This simplifies to:
\[ -2\sin^2(x) + a \sin x + (8 - 2a) = 0 \]

### Step 3: Form a quadratic equation
Now, we can express this as a standard quadratic equation in terms of \( \sin x \):
\[ -2y^2 + ay + (8 - 2a) = 0 \]
where \( y = \sin x \).

### Step 4: Apply the quadratic formula
For the quadratic equation \( Ay^2 + By + C = 0 \), the discriminant \( D \) must be non-negative for real solutions:
\[ D = B^2 - 4AC \]
In our case:
- \( A = -2 \)
- \( B = a \)
- \( C = 8 - 2a \)

Calculating the discriminant:
\[ D = a^2 - 4(-2)(8 - 2a) = a^2 + 8(8 - 2a) = a^2 + 64 - 16a \]
This simplifies to:
\[ D = a^2 - 16a + 64 \]

### Step 5: Set the discriminant greater than or equal to zero
For real solutions, we need:
\[ a^2 - 16a + 64 \geq 0 \]

### Step 6: Factor the quadratic
Factoring gives:
\[ (a - 8)^2 \geq 0 \]
This inequality holds for all \( a \) since a square is always non-negative.

### Step 7: Identify the critical point
The critical point occurs at \( a = 8 \). Thus, the quadratic is zero at this point and positive elsewhere.

### Step 8: Conclusion
The equation poses a real solution for all values of \( a \) except when \( a \) is exactly 8, where it has a double root. Therefore, the values of \( a \) for which the equation has real solutions are:
\[ a \leq 8 \quad \text{or} \quad a \geq 8 \]