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

Options

A. infinite number of 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 can follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the equation in a more manageable form. We can express \( e^{-\sin x} \) as \( \frac{1}{e^{\sin x}} \).

So, the equation becomes:
\[ e^{\sin x} - \frac{1}{e^{\sin x}} - 4 = 0 \]

### Step 2: Let \( A = e^{\sin x} \)
Next, we can substitute \( A \) for \( e^{\sin x} \). This gives us:
\[ A - \frac{1}{A} - 4 = 0 \]

### Step 3: Multiply through by \( A \)
To eliminate the fraction, we multiply the entire equation by \( A \) (assuming \( A \neq 0 \)):
\[ A^2 - 1 - 4A = 0 \]

### Step 4: Rearrange the equation
Rearranging the equation, we get:
\[ A^2 - 4A - 1 = 0 \]

### Step 5: Use the quadratic formula
Now we can use the quadratic formula to solve for \( A \):
\[ A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -4, c = -1 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-4)^2 - 4(1)(-1) = 16 + 4 = 20 \]

Now substituting into the quadratic formula:
\[ A = \frac{4 \pm \sqrt{20}}{2} \]
\[ A = \frac{4 \pm 2\sqrt{5}}{2} \]
\[ A = 2 \pm \sqrt{5} \]

### Step 6: Analyze the solutions for \( A \)
This gives us two possible values for \( A \):
1. \( A = 2 + \sqrt{5} \)
2. \( A = 2 - \sqrt{5} \)

### Step 7: Determine the range of \( A \)
Next, we need to evaluate whether these values are valid for \( A = e^{\sin x} \):
- \( 2 + \sqrt{5} \) is greater than 1 since \( \sqrt{5} \approx 2.236 \), thus \( 2 + \sqrt{5} \approx 4.236 \).
- \( 2 - \sqrt{5} \) is less than 0 since \( 2 - 2.236 < 0 \).

### Step 8: Conclusion about \( \sin x \)
Since \( A = e^{\sin x} \) must be positive (as the exponential function is always positive), the value \( 2 - \sqrt{5} \) is not a valid solution. Therefore, we only consider \( A = 2 + \sqrt{5} \).

Now, we find \( \sin x \):
\[ \sin x = \log_e(2 + \sqrt{5}) \]

### Step 9: Check the value of \( \sin x \)
We need to check if \( \log_e(2 + \sqrt{5}) \) falls within the range of \( \sin x \), which is \([-1, 1]\).

Calculating \( 2 + \sqrt{5} \):
- Since \( \sqrt{5} \approx 2.236 \), we have \( 2 + \sqrt{5} \approx 4.236 \).
- Therefore, \( \log_e(4.236) \) is greater than 1 (since \( e \approx 2.718 \)).

### Final Conclusion
Since \( \log_e(2 + \sqrt{5}) > 1 \) and the range of \( \sin x \) is \([-1, 1]\), there are no solutions for the equation \( e^{\sin x} - e^{-\sin x} - 4 = 0 \).

Thus, the equation has **no real roots**.

---