If for real values of x, `costheta=x+(1)/(x)`, then

Options

A. `theta` is an acute angle
B. `theta` is right angle
C. `theta` is an obtuse angle
D. No value of `theta` is possible

Explanation

To solve the problem where \( \cos \theta = x + \frac{1}{x} \), we need to analyze the expression and determine the possible values of \( \theta \).

### Step-by-Step Solution:

1. **Rewrite the equation**:
\[ \cos \theta = x + \frac{1}{x} \]

2. **Multiply through by \( x \)** (assuming \( x \neq 0 \)):
\[ x \cos \theta = x^2 + 1 \]

3. **Rearrange the equation**:
\[ x^2 - x \cos \theta + 1 = 0 \]
This is a quadratic equation in terms of \( x \).

4. **Identify coefficients**:
In the quadratic equation \( ax^2 + bx + c = 0 \), we have:
- \( a = 1 \)
- \( b = -\cos \theta \)
- \( c = 1 \)

5. **Calculate the discriminant**:
The discriminant \( D \) of a quadratic equation is given by:
\[ D = b^2 - 4ac \]
Substituting the values:
\[ D = (-\cos \theta)^2 - 4 \cdot 1 \cdot 1 = \cos^2 \theta - 4 \]

6. **Set the discriminant greater than or equal to zero for real values of \( x \)**:
For \( x \) to have real values, the discriminant must be non-negative:
\[ \cos^2 \theta - 4 \geq 0 \]

7. **Solve the inequality**:
Rearranging gives:
\[ \cos^2 \theta \geq 4 \]
Taking square roots (considering both positive and negative roots):
\[ \cos \theta \geq 2 \quad \text{or} \quad \cos \theta \leq -2 \]

8. **Analyze the range of \( \cos \theta \)**:
The cosine function has a range of \( [-1, 1] \). Therefore, \( \cos \theta \) cannot equal or exceed 2 or be less than -2.

9. **Conclusion**:
Since \( \cos \theta \) cannot take values outside of the interval \([-1, 1]\), the equation \( \cos \theta = x + \frac{1}{x} \) has no real solutions for \( \theta \).

Thus, the answer is:
**No value of \( \theta \) is possible.**