If `e^(sinx)-e^(-sinx)-4=0`, then the number of real values of x is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the equation \( e^{\sin x} - e^{-\sin x} - 4 = 0 \), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ e^{\sin x} - e^{-\sin x} - 4 = 0 \]
We can rewrite \( e^{-\sin x} \) as \( \frac{1}{e^{\sin x}} \):
\[ e^{\sin x} - \frac{1}{e^{\sin x}} - 4 = 0 \]

### Step 2: Multiply through by \( e^{\sin x} \)
To eliminate the fraction, we multiply through by \( e^{\sin x} \):
\[ (e^{\sin x})^2 - 1 - 4e^{\sin x} = 0 \]
This simplifies to:
\[ (e^{\sin x})^2 - 4e^{\sin x} - 1 = 0 \]

### Step 3: Substitute \( p = e^{\sin x} \)
Let \( p = e^{\sin x} \). Then the equation becomes:
\[ p^2 - 4p - 1 = 0 \]

### Step 4: Use the Quadratic Formula
We can solve this quadratic equation using the quadratic formula:
\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -4, c = -1 \):
\[ p = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ p = \frac{4 \pm \sqrt{16 + 4}}{2} \]
\[ p = \frac{4 \pm \sqrt{20}}{2} \]
\[ p = \frac{4 \pm 2\sqrt{5}}{2} \]
\[ p = 2 \pm \sqrt{5} \]

### Step 5: Determine the Values of \( p \)
Thus, we have two potential values for \( p \):
\[ p_1 = 2 + \sqrt{5}, \quad p_2 = 2 - \sqrt{5} \]

### Step 6: Evaluate the Range of \( p \)
Since \( p = e^{\sin x} \), we know that \( p \) must be positive. The range of \( e^{\sin x} \) is \( \left(\frac{1}{e}, e\right) \).

Now, we evaluate:
1. \( p_1 = 2 + \sqrt{5} \):
- Since \( \sqrt{5} \approx 2.236 \), then \( p_1 \approx 4.236 \), which is greater than \( e \) (approximately 2.718).

2. \( p_2 = 2 - \sqrt{5} \):
- Since \( \sqrt{5} \approx 2.236 \), then \( p_2 \approx -0.236 \), which is negative.

### Step 7: Conclusion
Neither \( p_1 \) nor \( p_2 \) lies within the range \( \left(\frac{1}{e}, e\right) \). Therefore, there are no real values of \( x \) that satisfy the original equation.

### Final Answer
The number of real values of \( x \) is:
\[ \boxed{0} \]