Find the values of p if it satisfy, `cos theta = x + p/x, x in R` for all real values of `theta`.

Answer

To find the values of \( p \) that satisfy the equation \( \cos \theta = x + \frac{p}{x} \) for all real values of \( \theta \), we can follow these steps:

### Step 1: Rearrange the equation
Starting from the equation:
\[ \cos \theta = x + \frac{p}{x} \]
we can multiply both sides by \( x \) (assuming \( x \neq 0 \)):
\[ x \cos \theta = x^2 + p \]

### Step 2: Form a quadratic equation
Rearranging the equation gives us:
\[ x^2 - x \cos \theta + p = 0 \]
This is a quadratic equation in terms of \( x \).

### Step 3: Apply the discriminant condition
For \( x \) to have real solutions, the discriminant of the quadratic equation must be greater than or equal to zero. The discriminant \( D \) for the equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
In our case:
- \( a = 1 \)
- \( b = -\cos \theta \)
- \( c = p \)

Thus, the discriminant becomes:
\[ D = (-\cos \theta)^2 - 4 \cdot 1 \cdot p = \cos^2 \theta - 4p \]
We require:
\[ \cos^2 \theta - 4p \geq 0 \]

### Step 4: Rearrange the inequality
Rearranging the inequality gives:
\[ \cos^2 \theta \geq 4p \]

### Step 5: Determine the range of \( \cos^2 \theta \)
The value of \( \cos^2 \theta \) ranges from \( 0 \) to \( 1 \) for all real values of \( \theta \). Therefore, we can write:
\[ 0 \leq \cos^2 \theta \leq 1 \]

### Step 6: Establish the condition for \( p \)
From the inequality \( \cos^2 \theta \geq 4p \), since the maximum value of \( \cos^2 \theta \) is \( 1 \), we have:
\[ 1 \geq 4p \]
This implies:
\[ p \leq \frac{1}{4} \]

### Conclusion
The values of \( p \) that satisfy the given equation for all real values of \( \theta \) are:
\[ p \leq \frac{1}{4} \]