The equation `"sin" theta = x +(p)/(x)` for real values of x is possible when

Options

A. `p gt 0`
B. `p le 0`
C. `p le (1)/(4)`
D. `p ge (1)/(2)`

Answer

To solve the equation \( \sin \theta = x + \frac{p}{x} \) for real values of \( x \), we can follow these steps:

### Step 1: Rearranging the Equation
Start by multiplying both sides of the equation by \( x \) (assuming \( x \neq 0 \)):
\[ x \sin \theta = x^2 + p \]
Rearranging gives us:
\[ x^2 - x \sin \theta + p = 0 \]

### Step 2: Discriminant Condition
For the quadratic equation \( x^2 - x \sin \theta + p = 0 \) to have real solutions for \( x \), the discriminant must be non-negative. The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = -\sin \theta \), and \( c = p \). Therefore, the discriminant is:
\[ D = (-\sin \theta)^2 - 4(1)(p) = \sin^2 \theta - 4p \]
We require:
\[ \sin^2 \theta - 4p \geq 0 \]

### Step 3: Solving the Inequality
Rearranging the inequality gives us:
\[ \sin^2 \theta \geq 4p \]
This implies:
\[ p \leq \frac{\sin^2 \theta}{4} \]

### Step 4: Considering the Range of \( \sin^2 \theta \)
Since \( \sin \theta \) can take values from -1 to 1, \( \sin^2 \theta \) will take values from 0 to 1. Therefore:
\[ 0 \leq \sin^2 \theta \leq 1 \]
This leads us to conclude:
\[ 0 \leq 4p \leq 1 \]
Thus, we can divide by 4:
\[ 0 \leq p \leq \frac{1}{4} \]

### Conclusion
The possible values of \( p \) for which the equation \( \sin \theta = x + \frac{p}{x} \) has real solutions for \( x \) are:
\[ p \in [0, \frac{1}{4}] \]

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