The solution set of `"sin" (x - (pi)/(4))-"cos" (x + (3 pi)/(4)) =1 " and " (2 "cos" 7x)/("cos" 3 + "sin" 3) gt 2^("cos" 2x)`, is

Options

A. `k pi + (-1)^(k) (pi)/(4), k in Z`
B. `(8k + 3) (pi)/(4), k in Z`
C. `(8k + 1) (pi)/(4), k in Z`
D. none of these

Explanation

To solve the given problem, we need to tackle the two equations separately and then find the common solution set.

### Step 1: Solve the first equation
The first equation is:
\[ \sin\left(x - \frac{\pi}{4}\right) - \cos\left(x + \frac{3\pi}{4}\right) = 1 \]

Using the cosine angle addition formula, we can rewrite \(\cos\left(x + \frac{3\pi}{4}\right)\) as:
\[ \cos\left(x + \frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x \]

Thus, the equation becomes:
\[ \sin\left(x - \frac{\pi}{4}\right) + \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = 1 \]

Using the sine subtraction formula:
\[ \sin\left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}(\sin x - \cos x) \]

Substituting this into the equation gives:
\[ \frac{1}{\sqrt{2}}(\sin x - \cos x) + \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = 1 \]

Combining terms:
\[ \frac{1}{\sqrt{2}}(2\sin x) = 1 \]
\[ \sin x = \frac{\sqrt{2}}{2} \]

### Step 2: Find the general solution for \(\sin x = \frac{\sqrt{2}}{2}\)
The solutions for \(\sin x = \frac{\sqrt{2}}{2}\) are:
\[ x = n\pi + \frac{\pi}{4} \quad \text{or} \quad x = n\pi + \frac{3\pi}{4}, \quad n \in \mathbb{Z} \]

### Step 3: Solve the second inequality
The second inequality is:
\[ \frac{2 \cos(7x)}{\cos 3 + \sin 3} > 2^{\cos(2x)} \]

Since \(\cos 3 + \sin 3\) is a constant, we can denote it as \(k\), where \(k = \cos 3 + \sin 3\).

Rearranging gives:
\[ \cos(7x) > \frac{k}{2} \cdot 2^{\cos(2x)} \]

### Step 4: Analyze \(\cos(2x)\)
Since \(\cos(2x)\) can take values from -1 to 1, we need to analyze the inequality further.

1. If \(\cos(2x) = 0\), then \(2^{\cos(2x)} = 1\).
2. If \(\cos(2x) = 1\), then \(2^{\cos(2x)} = 2\).
3. If \(\cos(2x) = -1\), then \(2^{\cos(2x)} = \frac{1}{2}\).

### Step 5: Combine the results
From the first equation, we have:
\[ x = n\pi + \frac{\pi}{4} \quad \text{or} \quad x = n\pi + \frac{3\pi}{4} \]

From the second inequality, we need to find values of \(x\) such that \(\cos(7x) > \frac{k}{2} \cdot 2^{\cos(2x)}\) holds true.

### Final Solution Set
The solution set of the original problem is:
\[ x = 2n\pi + \frac{3\pi}{4}, \quad n \in \mathbb{Z} \]