Solve the following system of equations. ` sin x + cos y=1, cos 2x-cos 2y=1 `

Answer

To solve the system of equations given by:

1. \( \sin x + \cos y = 1 \)
2. \( \cos 2x - \cos 2y = 1 \)

we will follow these steps:

### Step 1: Analyze the first equation
From the first equation, we can express \( \cos y \) in terms of \( \sin x \):

\[ \cos y = 1 - \sin x \]

### Step 2: Use the identity for cosine
We know the double angle identity for cosine:

\[ \cos 2x = 2\cos^2 x - 1 \quad \text{and} \quad \cos 2y = 2\cos^2 y - 1 \]

### Step 3: Substitute the identities into the second equation
Substituting these identities into the second equation, we have:

\[ (2\cos^2 x - 1) - (2\cos^2 y - 1) = 1 \]

This simplifies to:

\[ 2\cos^2 x - 2\cos^2 y = 1 \]

### Step 4: Rearranging the equation
Rearranging gives:

\[ \cos^2 x - \cos^2 y = \frac{1}{2} \]

### Step 5: Substitute \( \cos y \)
Using \( \cos y = 1 - \sin x \), we can express \( \cos^2 y \):

\[ \cos^2 y = (1 - \sin x)^2 = 1 - 2\sin x + \sin^2 x \]

### Step 6: Substitute into the equation
Now substitute \( \cos^2 y \) into the equation:

\[ \cos^2 x - (1 - 2\sin x + \sin^2 x) = \frac{1}{2} \]

### Step 7: Simplify the equation
This leads to:

\[ \cos^2 x - 1 + 2\sin x - \sin^2 x = \frac{1}{2} \]

### Step 8: Use the identity \( \cos^2 x + \sin^2 x = 1 \)
Using the identity \( \cos^2 x + \sin^2 x = 1 \):

\[ \cos^2 x = 1 - \sin^2 x \]

Substituting this in gives:

\[ (1 - \sin^2 x) - 1 + 2\sin x - \sin^2 x = \frac{1}{2} \]

### Step 9: Combine like terms
This simplifies to:

\[ -2\sin^2 x + 2\sin x = \frac{1}{2} \]

### Step 10: Rearranging the quadratic equation
Multiplying through by -1 and rearranging gives:

\[ 2\sin^2 x - 2\sin x + \frac{1}{2} = 0 \]

### Step 11: Multiply through by 2 to eliminate the fraction
This results in:

\[ 4\sin^2 x - 4\sin x + 1 = 0 \]

### Step 12: Solve the quadratic equation
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 4, b = -4, c = 1 \):

\[ \sin x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} \]

Calculating the discriminant:

\[ \sin x = \frac{4 \pm \sqrt{16 - 16}}{8} = \frac{4 \pm 0}{8} = \frac{1}{2} \]

### Step 13: Find the values of \( x \)
Thus, \( \sin x = \frac{1}{2} \) implies:

\[ x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi \]

### Step 14: Find \( \cos y \)
Substituting \( \sin x = \frac{1}{2} \) into \( \cos y = 1 - \sin x \):

\[ \cos y = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 15: Find the values of \( y \)
Thus, \( \cos y = \frac{1}{2} \) implies:

\[ y = \frac{\pi}{3} + 2m\pi \quad \text{or} \quad y = -\frac{\pi}{3} + 2m\pi \]

### Final Solution
The solution set for the system of equations is:

\[ x = \frac{\pi}{6} + 2n\pi, \quad x = \frac{5\pi}{6} + 2n\pi \]
\[ y = \frac{\pi}{3} + 2m\pi, \quad y = -\frac{\pi}{3} + 2m\pi \]