If x and `alpha` are real, then the inequation
`log_(2)x+log_(x)2+2cos alpha le 0`

Options

A. has no solution
B. has exactly two solutions
C. is satisfied for any real `alpha` and any real x in (0,1)
D. None of these

Answer

To solve the inequation \( \log_2 x + \log_x 2 + 2 \cos \alpha \leq 0 \), we can follow these steps:

### Step 1: Rewrite the logarithmic terms
We start with the given inequation:
\[ \log_2 x + \log_x 2 + 2 \cos \alpha \leq 0 \]
Using the change of base formula, we can rewrite \( \log_x 2 \) as \( \frac{1}{\log_2 x} \). Let \( t = \log_2 x \). Thus, we have:
\[ t + \frac{1}{t} + 2 \cos \alpha \leq 0 \]

### Step 2: Multiply through by \( t \)
To eliminate the fraction, we multiply the entire inequation by \( t \) (noting that \( t > 0 \) since \( x > 0 \)):
\[ t^2 + 1 + 2t \cos \alpha \leq 0 \]

### Step 3: Rearrange into a standard quadratic form
Rearranging gives us:
\[ t^2 + 2t \cos \alpha + 1 \leq 0 \]

### Step 4: Identify the coefficients
Here, we can identify:
- \( a = 1 \)
- \( b = 2 \cos \alpha \)
- \( c = 1 \)

### Step 5: Calculate the discriminant
The discriminant \( D \) of the quadratic equation \( at^2 + bt + c = 0 \) is given by:
\[ D = b^2 - 4ac = (2 \cos \alpha)^2 - 4 \cdot 1 \cdot 1 = 4 \cos^2 \alpha - 4 \]
This simplifies to:
\[ D = 4(\cos^2 \alpha - 1) = 4(-\sin^2 \alpha) \]

### Step 6: Analyze the discriminant
For the quadratic to have real roots, the discriminant must be non-negative:
\[ 4(-\sin^2 \alpha) \geq 0 \]
This implies:
\[ -\sin^2 \alpha \geq 0 \implies \sin^2 \alpha = 0 \]
Thus, \( \sin \alpha = 0 \), which occurs when:
\[ \alpha = n\pi \quad (n \in \mathbb{Z}) \]

### Step 7: Solve for \( t \)
If \( \sin \alpha = 0 \), then \( \cos \alpha = \pm 1 \). We can substitute \( \cos \alpha = 1 \) into the quadratic:
\[ t^2 + 2t + 1 \leq 0 \]
This factors to:
\[ (t + 1)^2 \leq 0 \]
The only solution is:
\[ t + 1 = 0 \implies t = -1 \]

### Step 8: Convert back to \( x \)
Recalling that \( t = \log_2 x \):
\[ \log_2 x = -1 \implies x = 2^{-1} = \frac{1}{2} \]

### Step 9: Determine the range of \( x \)
Since \( t = \log_2 x \) must be positive (as \( x > 0 \)), we need to check the conditions:
- \( x \) must be in the range \( (0, 1) \) since \( \log_2 x \) is negative for \( 0 < x < 1 \).

### Step 10: Conclusion
Thus, the solution to the inequation is:
\[ x \in \left( \frac{1}{2}, 2 \right) \]