The equation `e^(sin x)-e^(-sin x)-4=0` has

Options

A. Infinite number of real roots
B. no real roots
C. exactly one real root
D. exactly four real roots.

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 given equation:
\[ e^{\sin x} - e^{-\sin x} - 4 = 0 \]
We can rewrite it as:
\[ e^{\sin x} - \frac{1}{e^{\sin x}} - 4 = 0 \]

### Step 2: Substitute \( t \)
Let \( t = e^{\sin x} \). Then, we can rewrite the equation as:
\[ t - \frac{1}{t} - 4 = 0 \]
Multiplying through by \( t \) (assuming \( t \neq 0 \)):
\[ t^2 - 1 - 4t = 0 \]

### Step 3: Rearrange the equation
Rearranging gives us a standard quadratic equation:
\[ t^2 - 4t - 1 = 0 \]

### Step 4: Apply the quadratic formula
We will use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -4, c = -1 \):
\[ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ t = \frac{4 \pm \sqrt{16 + 4}}{2} \]
\[ t = \frac{4 \pm \sqrt{20}}{2} \]
\[ t = \frac{4 \pm 2\sqrt{5}}{2} \]
\[ t = 2 \pm \sqrt{5} \]

### Step 5: Determine valid solutions for \( t \)
Since \( t = e^{\sin x} \) must be positive, we only consider:
\[ t = 2 + \sqrt{5} \]
We discard \( t = 2 - \sqrt{5} \) because \( 2 - \sqrt{5} < 0 \).

### Step 6: Solve for \( \sin x \)
Now, we substitute back to find \( \sin x \):
\[ e^{\sin x} = 2 + \sqrt{5} \]
Taking the natural logarithm of both sides:
\[ \sin x = \ln(2 + \sqrt{5}) \]

### Step 7: Analyze the value of \( \sin x \)
Next, we need to check if \( \ln(2 + \sqrt{5}) \) is within the range of the sine function, which is \([-1, 1]\).

Calculating \( 2 + \sqrt{5} \):
\[ \sqrt{5} \approx 2.236 \implies 2 + \sqrt{5} \approx 4.236 \]
Thus:
\[ \ln(2 + \sqrt{5}) \approx \ln(4.236) \approx 1.446 \]
Since \( \ln(2 + \sqrt{5}) > 1 \), it is not possible for \( \sin x \) to take this value.

### Conclusion
Since \( \sin x \) cannot be greater than 1, the equation \( e^{\sin x} - e^{-\sin x} - 4 = 0 \) has no real solutions.

### Final Answer
The equation has **no real roots**.

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