The general solution of `sec 4x -sec 2x =2` is

Options

A. `((2n+1)pi)/(10),((2m+1)pi)/(2), n, m in l`
B. `((2n+1)pi)/(5),((2m+1)pi)/(5), n, m in l`
C. `((2n+2)pi)/(10),((2m+4)pi)/(2), n, m in l`
D. None of these

Answer

To solve the equation \( \sec 4x - \sec 2x = 2 \), we will follow these steps:

### Step 1: Rewrite the secant functions
We start by rewriting the secant functions in terms of cosine:
\[ \sec 4x = \frac{1}{\cos 4x} \quad \text{and} \quad \sec 2x = \frac{1}{\cos 2x} \]
Thus, the equation becomes:
\[ \frac{1}{\cos 4x} - \frac{1}{\cos 2x} = 2 \]

### Step 2: Find a common denominator
To combine the fractions, we find a common denominator:
\[ \frac{\cos 2x - \cos 4x}{\cos 4x \cos 2x} = 2 \]
Multiplying both sides by \( \cos 4x \cos 2x \) gives:
\[ \cos 2x - \cos 4x = 2 \cos 4x \cos 2x \]

### Step 3: Rearranging the equation
Rearranging the equation leads to:
\[ \cos 2x - \cos 4x - 2 \cos 4x \cos 2x = 0 \]
This can be rewritten as:
\[ \cos 2x - \cos 4x (1 + 2 \cos 2x) = 0 \]

### Step 4: Factor the equation
We can factor this equation:
\[ \cos 2x - \cos 4x (1 + 2 \cos 2x) = 0 \]
This implies:
\[ \cos 4x (1 + 2 \cos 2x) = \cos 2x \]

### Step 5: Use the cosine subtraction formula
Using the cosine subtraction formula, we can express \( \cos 4x \) in terms of \( \cos 2x \):
\[ \cos 4x = 2 \cos^2 2x - 1 \]
Substituting this back into the equation gives:
\[ \cos 2x - (2 \cos^2 2x - 1)(1 + 2 \cos 2x) = 0 \]

### Step 6: Solve for \( \cos 2x \)
This leads to a polynomial equation in \( \cos 2x \). Solving this polynomial will yield the values for \( \cos 2x \).

### Step 7: Find the general solutions
Once we have the values for \( \cos 2x \), we can find the general solutions for \( x \):
1. If \( \cos 2x = 0 \), then \( 2x = \frac{\pi}{2} + n\pi \) leading to \( x = \frac{\pi}{4} + \frac{n\pi}{2} \).
2. If \( \cos 2x = k \) for some \( k \), then \( 2x = \pm \cos^{-1}(k) + 2n\pi \) leading to \( x = \frac{1}{2} \cos^{-1}(k) + n\pi \).

### Final General Solution
Combining all solutions, we find:
\[ x = \frac{2n + 1}{10} \pi \quad \text{and} \quad x = \frac{2m + 1}{2} \pi \]
for integers \( n \) and \( m \).