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 start by rearranging the equation into a standard quadratic form.

### Step 1: Rearranging the Equation
We can rewrite the equation as:
\[ \sin \theta = x + \frac{p}{x} \]
Multiplying both sides by \( x \) (assuming \( x \neq 0 \)):
\[ x \sin \theta = x^2 + p \]
Rearranging gives us:
\[ x^2 - x \sin \theta + p = 0 \]

### Step 2: Identifying the Quadratic Coefficients
In the quadratic equation \( ax^2 + bx + c = 0 \), we identify:
- \( a = 1 \)
- \( b = -\sin \theta \)
- \( c = p \)

### Step 3: Applying the Discriminant Condition
For the quadratic equation to have real solutions, the discriminant \( D \) must be greater than or equal to zero:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ (-\sin \theta)^2 - 4(1)(p) \geq 0 \]
This simplifies to:
\[ \sin^2 \theta - 4p \geq 0 \]

### Step 4: Rearranging the Inequality
Rearranging the inequality gives:
\[ \sin^2 \theta \geq 4p \]
Thus, we can express \( p \) in terms of \( \sin^2 \theta \):
\[ p \leq \frac{\sin^2 \theta}{4} \]

### Step 5: Finding the Range of \( \sin^2 \theta \)
The sine function \( \sin \theta \) can take values between -1 and 1. Therefore, \( \sin^2 \theta \) will range from:
\[ 0 \leq \sin^2 \theta \leq 1 \]

### Step 6: Determining the Range of \( p \)
From the inequality \( p \leq \frac{\sin^2 \theta}{4} \), we can find the maximum value of \( p \):
- When \( \sin^2 \theta = 1 \), we have:
\[ p \leq \frac{1}{4} \]
- When \( \sin^2 \theta = 0 \), \( p \) can be any non-negative value, but since we are looking for the maximum, we focus on the upper limit.

### Conclusion
Thus, the condition for \( p \) for the equation \( \sin \theta = x + \frac{p}{x} \) to have real solutions is:
\[ p \leq \frac{1}{4} \]