If equation `x^2+2x+2 + e^alpha - 2 sin beta = 0` has a real solution in `x,` then (where `n in ZZ`)

Options

A. `alpha, beta in R`
B. `alpha in(0,1), beta in(pi//6,pi//2)`
C. `alpha in(0,1),beta in(2n pi+pi//6,2n pi+5pi//6)`
D. `alpha in (-oo,0],beta in (2n pi-pi//6)`

Explanation

To solve the equation \( x^2 + 2x + 2 + e^{\alpha} - 2\sin(\beta) = 0 \) for real solutions in \( x \), we need to analyze the conditions under which this quadratic equation has real roots.

### Step 1: Identify the quadratic expression
The given equation can be rewritten as:
\[ x^2 + 2x + (2 + e^{\alpha} - 2\sin(\beta)) = 0 \]
This is a standard quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = 2 \), and \( c = 2 + e^{\alpha} - 2\sin(\beta) \).

### Step 2: Apply the discriminant condition
For the quadratic equation to have real solutions, the discriminant must be non-negative. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = 2^2 - 4 \cdot 1 \cdot (2 + e^{\alpha} - 2\sin(\beta)) \]
\[ D = 4 - 4(2 + e^{\alpha} - 2\sin(\beta)) \]
\[ D = 4 - 8 - 4e^{\alpha} + 8\sin(\beta) \]
\[ D = -4e^{\alpha} + 8\sin(\beta) - 4 \]

### Step 3: Set the discriminant greater than or equal to zero
To ensure real solutions, we set the discriminant \( D \) to be greater than or equal to zero:
\[ -4e^{\alpha} + 8\sin(\beta) - 4 \geq 0 \]
Rearranging gives:
\[ 8\sin(\beta) \geq 4 + 4e^{\alpha} \]
Dividing through by 4:
\[ 2\sin(\beta) \geq 1 + e^{\alpha} \]
Thus, we have:
\[ e^{\alpha} \leq 2\sin(\beta) - 1 \]

### Step 4: Analyze the sine function
The sine function \( \sin(\beta) \) has a range of \([-1, 1]\). Therefore, we need to find the range of \( 2\sin(\beta) - 1 \):
- When \( \sin(\beta) = -1 \):
\[ 2(-1) - 1 = -3 \]
- When \( \sin(\beta) = 1 \):
\[ 2(1) - 1 = 1 \]

Thus, the expression \( 2\sin(\beta) - 1 \) ranges from \(-3\) to \(1\). Therefore, we have:
\[ e^{\alpha} \leq 1 \]

### Step 5: Solve for \( \alpha \)
Since \( e^{\alpha} \) is always positive, we can conclude:
\[ e^{\alpha} \leq 1 \implies \alpha \leq 0 \]

### Step 6: Determine the interval for \( \beta \)
Now we need to find the interval for \( \beta \) such that \( 2\sin(\beta) \geq 1 + e^{\alpha} \). The maximum value of \( e^{\alpha} \) is \( 1 \) when \( \alpha = 0 \):
\[ 2\sin(\beta) \geq 1 + 1 = 2 \implies \sin(\beta) \geq 1 \]
This means \( \beta = \frac{\pi}{2} + 2n\pi \) for \( n \in \mathbb{Z} \).

### Final Result
The intervals for \( \alpha \) and \( \beta \) are:
- \( \alpha \leq 0 \) (or \( (-\infty, 0] \))
- \( \beta = \frac{\pi}{2} + 2n\pi \) for \( n \in \mathbb{Z} \)