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 values of \( \alpha \) that allow at least one solution, we will follow these steps:

### Step 1: Simplify the equation
We start with the equation:
\[ \cos^4 x - \sin^4 x + \cos 2x + \alpha^2 + \alpha = 0 \]
We can use the identity \( a^2 - b^2 = (a-b)(a+b) \) to rewrite \( \cos^4 x - \sin^4 x \):
\[ \cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) \]
Since \( \cos^2 x + \sin^2 x = 1 \), this simplifies to:
\[ \cos^4 x - \sin^4 x = \cos^2 x - \sin^2 x = \cos 2x \]
Thus, we can rewrite the original equation as:
\[ \cos 2x + \cos 2x + \alpha^2 + \alpha = 0 \]
This simplifies to:
\[ 2\cos 2x + \alpha^2 + \alpha = 0 \]

### Step 2: Rearranging the equation
Rearranging gives:
\[ 2\cos 2x = -(\alpha^2 + \alpha) \]

### Step 3: Analyze the range of \( \cos 2x \)
The cosine function has a range of \([-1, 1]\). Therefore, we need:
\[ -1 \leq 2\cos 2x \leq 1 \]
This leads to the inequalities:
\[ -1 \leq -(\alpha^2 + \alpha) \leq 1 \]

### Step 4: Solve the inequalities
1. From \( -1 \leq -(\alpha^2 + \alpha) \):
\[ \alpha^2 + \alpha \leq 1 \]
Rearranging gives:
\[ \alpha^2 + \alpha - 1 \leq 0 \]
The roots of the quadratic equation \( \alpha^2 + \alpha - 1 = 0 \) can be found using the quadratic formula:
\[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]
The roots are:
\[ \alpha_1 = \frac{-1 - \sqrt{5}}{2}, \quad \alpha_2 = \frac{-1 + \sqrt{5}}{2} \]
The quadratic opens upwards (since the coefficient of \( \alpha^2 \) is positive), so it is less than or equal to zero between the roots:
\[ \frac{-1 - \sqrt{5}}{2} \leq \alpha \leq \frac{-1 + \sqrt{5}}{2} \]

2. From \( -(\alpha^2 + \alpha) \leq 1 \):
\[ \alpha^2 + \alpha + 1 \geq 0 \]
The discriminant of this quadratic is:
\[ D = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative, this quadratic is always positive for all \( \alpha \).

### Step 5: Combine the results
The only condition we need to satisfy is:
\[ \frac{-1 - \sqrt{5}}{2} \leq \alpha \leq \frac{-1 + \sqrt{5}}{2} \]

### Conclusion
The equation will have at least one solution if:
\[ \alpha \in \left[\frac{-1 - \sqrt{5}}{2}, \frac{-1 + \sqrt{5}}{2}\right] \]