Find the range of y such that the equation in x , ` y + cos x = sin x ` has a real solutions . For `y=1` , find x such that ` 0 lt x lt 2 pi`

Answer

To solve the problem, we need to find the range of \( y \) such that the equation \( y + \cos x = \sin x \) has real solutions. Then, for \( y = 1 \), we will find \( x \) such that \( 0 < x < 2\pi \).

### Step 1: Rearranging the Equation
We start with the equation:
\[ y + \cos x = \sin x \]
Rearranging gives us:
\[ y = \sin x - \cos x \]

### Step 2: Finding the Range of \( \sin x - \cos x \)
To find the range of \( \sin x - \cos x \), we can express it in a different form. We can use the identity:
\[ \sin x - \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x \right) \]
This can be rewritten as:
\[ \sin x - \cos x = \sqrt{2} \sin \left(x - \frac{\pi}{4}\right) \]

### Step 3: Determining the Range
The sine function oscillates between -1 and 1, so:
\[ -\sqrt{2} \leq \sqrt{2} \sin \left(x - \frac{\pi}{4}\right) \leq \sqrt{2} \]
Thus, the range of \( y \) is:
\[ -\sqrt{2} \leq y \leq \sqrt{2} \]

### Step 4: Finding \( x \) for \( y = 1 \)
Now, we set \( y = 1 \):
\[ 1 = \sin x - \cos x \]
Rearranging gives:
\[ \sin x - \cos x = 1 \]
Using the same transformation as before:
\[ \sqrt{2} \sin \left(x - \frac{\pi}{4}\right) = 1 \]
Dividing both sides by \( \sqrt{2} \):
\[ \sin \left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Step 5: Solving for \( x \)
The general solution for \( \sin \theta = \frac{1}{\sqrt{2}} \) is:
\[ \theta = \frac{\pi}{4} + 2n\pi \quad \text{or} \quad \theta = \frac{3\pi}{4} + 2n\pi \]
Substituting back for \( x \):
1. \( x - \frac{\pi}{4} = \frac{\pi}{4} + 2n\pi \) leads to:
\[ x = \frac{\pi}{2} + 2n\pi \]
2. \( x - \frac{\pi}{4} = \frac{3\pi}{4} + 2n\pi \) leads to:
\[ x = \pi + 2n\pi \]

### Step 6: Finding Values in \( (0, 2\pi) \)
For \( n = 0 \):
1. From \( x = \frac{\pi}{2} \), we have \( x = \frac{\pi}{2} \).
2. From \( x = \pi \), we have \( x = \pi \).

Both values \( \frac{\pi}{2} \) and \( \pi \) are within the interval \( (0, 2\pi) \).

### Final Answer
The range of \( y \) for which the equation has real solutions is:
\[ [-\sqrt{2}, \sqrt{2}] \]
For \( y = 1 \), the values of \( x \) in the interval \( (0, 2\pi) \) are:
\[ x = \frac{\pi}{2}, \quad x = \pi \]