The equation `cos^(4)x-sin^(4)x+cos2x+alpha^(2)+alpha=0` will have at least one solution, if

Options

A. `-2 le alpha le 2`
B. `-3le alpha le 1`
C. `-2 le alpha le 1`
D. `-1 le alpha le2`

Answer

To solve the equation \( \cos^4 x - \sin^4 x + \cos 2x + \alpha^2 + \alpha = 0 \) for the condition that it has at least one solution, we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos^4 x - \sin^4 x + \cos 2x + \alpha^2 + \alpha = 0 \]

Using the identity \( a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) \) and knowing that \( \cos^2 x + \sin^2 x = 1 \), we can rewrite \( \cos^4 x - \sin^4 x \) as:
\[ \cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = \cos 2x \]
Thus, we can simplify the equation to:
\[ \cos 2x + \cos 2x + \alpha^2 + \alpha = 0 \]
This simplifies to:
\[ 2\cos 2x + \alpha^2 + \alpha = 0 \]

### Step 2: Isolate \( \cos 2x \)
Rearranging gives:
\[ 2\cos 2x = -(\alpha^2 + \alpha) \]
Dividing by 2:
\[ \cos 2x = -\frac{1}{2}(\alpha^2 + \alpha) \]

### Step 3: Determine the range of \( \cos 2x \)
The cosine function has a range of \([-1, 1]\). Therefore, for the equation to have at least one solution, we need:
\[ -1 \leq -\frac{1}{2}(\alpha^2 + \alpha) \leq 1 \]

### Step 4: Solve the inequalities
1. **First Inequality**:
\[ -1 \leq -\frac{1}{2}(\alpha^2 + \alpha) \]
Multiplying through by -2 (and reversing the inequality):
\[ 2 \geq \alpha^2 + \alpha \]
Rearranging gives:
\[ \alpha^2 + \alpha - 2 \leq 0 \]
Factoring:
\[ (\alpha - 1)(\alpha + 2) \leq 0 \]
The solution to this inequality is:
\[ -2 \leq \alpha \leq 1 \]

2. **Second Inequality**:
\[ -\frac{1}{2}(\alpha^2 + \alpha) \leq 1 \]
Multiplying through by -2 (and reversing the inequality):
\[ \alpha^2 + \alpha \geq -2 \]
Rearranging gives:
\[ \alpha^2 + \alpha + 2 \geq 0 \]
This quadratic is always non-negative for all real values of \( \alpha \) because its discriminant is negative:
\[ D = 1^2 - 4 \cdot 1 \cdot 2 = 1 - 8 = -7 < 0 \]

### Step 5: Combine results
From the first inequality, we have:
\[ -2 \leq \alpha \leq 1 \]
The second inequality does not impose any additional restrictions. Therefore, the final solution is:
\[ \alpha \in [-2, 1] \]

### Final Answer
The equation will have at least one solution if:
\[ -2 \leq \alpha \leq 1 \]