Solve the equation for x and y , `|sin x + cos x|^( sin^(2) x-1//4)=1+|siny| and cos^(2)y=1+sin^(2)y`.

Answer

To solve the equations given in the problem, we will break it down into two parts: solving for \(y\) first, and then solving for \(x\).

### Step 1: Solve the second equation for \(y\)

The second equation is:
\[ \cos^2 y = 1 + \sin^2 y \]

Using the Pythagorean identity, we know that:
\[ \cos^2 y + \sin^2 y = 1 \]

We can substitute \(\cos^2 y\) from the identity into the equation:
\[ 1 - \sin^2 y = 1 + \sin^2 y \]

Now, simplifying this gives:
\[ - \sin^2 y = \sin^2 y \]

Adding \(\sin^2 y\) to both sides results in:
\[ 0 = 2\sin^2 y \]

This implies:
\[ \sin^2 y = 0 \]

Taking the square root of both sides, we find:
\[ \sin y = 0 \]

The general solution for \(\sin y = 0\) is:
\[ y = n\pi \quad (n \in \mathbb{Z}) \]

### Step 2: Solve the first equation for \(x\)

The first equation is:
\[ |\sin x + \cos x|^{\sin^2 x - \frac{1}{4}} = 1 + |\sin y| \]

Since we found \(y = n\pi\), we know:
\[ \sin y = 0 \quad \Rightarrow \quad 1 + |\sin y| = 1 + 0 = 1 \]

Thus, we can rewrite the first equation as:
\[ |\sin x + \cos x|^{\sin^2 x - \frac{1}{4}} = 1 \]

For the expression to equal 1, we have two cases to consider:

1. **Case 1:** \( |\sin x + \cos x| = 1 \)
2. **Case 2:** \( \sin^2 x - \frac{1}{4} = 0 \)

#### Case 1: \( |\sin x + \cos x| = 1 \)

This occurs when:
\[ \sin x + \cos x = 1 \quad \text{or} \quad \sin x + \cos x = -1 \]

For \(\sin x + \cos x = 1\):
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) = 1 \]
This leads to:
\[ \sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad x + \frac{\pi}{4} = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad x + \frac{\pi}{4} = \frac{3\pi}{4} + 2k\pi \]
Solving gives:
\[ x = 2k\pi \quad \text{or} \quad x = \frac{\pi}{2} + 2k\pi \]

For \(\sin x + \cos x = -1\):
\[ \sin x + \cos x = -\sqrt{2} \sin\left(x + \frac{\pi}{4}\right) = -1 \]
This leads to:
\[ \sin\left(x + \frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} \quad \Rightarrow \quad x + \frac{\pi}{4} = \frac{7\pi}{4} + 2k\pi \quad \Rightarrow \quad x = \frac{3\pi}{2} + 2k\pi \]

#### Case 2: \( \sin^2 x - \frac{1}{4} = 0 \)

This gives:
\[ \sin^2 x = \frac{1}{4} \quad \Rightarrow \quad \sin x = \frac{1}{2} \quad \text{or} \quad \sin x = -\frac{1}{2} \]

For \(\sin x = \frac{1}{2}\):
\[ x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2k\pi \]

For \(\sin x = -\frac{1}{2}\):
\[ x = \frac{7\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{11\pi}{6} + 2k\pi \]

### Final Solutions

Combining all the solutions for \(x\) and \(y\):
- \(y = n\pi\) where \(n \in \mathbb{Z}\)
- \(x = 2k\pi\), \(x = \frac{\pi}{2} + 2k\pi\), \(x = \frac{3\pi}{2} + 2k\pi\), \(x = \frac{\pi}{6} + 2k\pi\), \(x = \frac{5\pi}{6} + 2k\pi\), \(x = \frac{7\pi}{6} + 2k\pi\), \(x = \frac{11\pi}{6} + 2k\pi\) where \(k \in \mathbb{Z}\).