If `e^(cos x) -e^(-cos x) = 4`, then the value of cos x, is

Options

A. `log_(e)(2 + sqrt(5))`
B. `-log_(e)(2 + sqrt(5))`
C. `log_(e)(-2 + sqrt(5))`
D. none of these

Answer

To solve the equation \( e^{\cos x} - e^{-\cos x} = 4 \) for the value of \( \cos x \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ e^{\cos x} - e^{-\cos x} = 4 \]
Let \( t = \cos x \). Then, we can rewrite the equation as:
\[ e^t - e^{-t} = 4 \]

### Step 2: Multiply by \( e^t \)
To eliminate the negative exponent, multiply both sides by \( e^t \):
\[ e^{2t} - 1 = 4e^t \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ e^{2t} - 4e^t - 1 = 0 \]

### Step 4: Substitute \( y = e^t \)
Let \( y = e^t \). Then, we can rewrite the equation as:
\[ y^2 - 4y - 1 = 0 \]

### Step 5: Solve the quadratic equation
Now we can use the quadratic formula to solve for \( y \):
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, \( a = 1 \), \( b = -4 \), and \( c = -1 \):
\[ y = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ y = \frac{4 \pm \sqrt{16 + 4}}{2} \]
\[ y = \frac{4 \pm \sqrt{20}}{2} \]
\[ y = \frac{4 \pm 2\sqrt{5}}{2} \]
\[ y = 2 \pm \sqrt{5} \]

### Step 6: Determine valid solutions for \( y \)
Since \( y = e^t \) must be positive, we discard \( 2 - \sqrt{5} \) because \( \sqrt{5} \) is approximately \( 2.236 \), making \( 2 - \sqrt{5} \) negative. Thus, we have:
\[ y = 2 + \sqrt{5} \]

### Step 7: Find \( t \)
Now, we can find \( t \):
\[ e^t = 2 + \sqrt{5} \]
Taking the natural logarithm of both sides:
\[ t = \ln(2 + \sqrt{5}) \]

### Step 8: Substitute back to find \( \cos x \)
Since \( t = \cos x \), we get:
\[ \cos x = \ln(2 + \sqrt{5}) \]

### Conclusion
Thus, the value of \( \cos x \) is:
\[ \cos x = \ln(2 + \sqrt{5}) \]