Find the intervals in which `f(x)=sinx-cosx`, where `0ltxlt2pi`, is strictly increasing or decreasing.

Answer

To solve the problem of finding the intervals in which the function \( f(x) = \sin x - \cos x \) is strictly increasing or decreasing for \( 0 < x < 2\pi \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \):

\[ f'(x) = \frac{d}{dx}(\sin x - \cos x) = \cos x + \sin x \]

### Step 2: Set the derivative equal to zero to find critical points
Next, we find the critical points by setting the derivative equal to zero:

\[ \cos x + \sin x = 0 \]

### Step 3: Solve for \( x \)
We can rearrange the equation:

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

Dividing both sides by \( \cos x \) (where \( \cos x \neq 0 \)) gives:

\[ \tan x = -1 \]

The general solutions for \( \tan x = -1 \) are:

\[ x = \frac{3\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

Within the interval \( 0 < x < 2\pi \), the specific solutions are:

1. \( x = \frac{3\pi}{4} \)
2. \( x = \frac{7\pi}{4} \)

### Step 4: Determine the sign of the derivative in the intervals
Now we will check the sign of \( f'(x) \) in the intervals determined by the critical points:

1. Interval \( (0, \frac{3\pi}{4}) \)
2. Interval \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \)
3. Interval \( (\frac{7\pi}{4}, 2\pi) \)

#### Check the sign in \( (0, \frac{3\pi}{4}) \)
Choose a test point, e.g., \( x = \frac{\pi}{4} \):

\[ f'(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} > 0 \]

Thus, \( f(x) \) is **increasing** in \( (0, \frac{3\pi}{4}) \).

#### Check the sign in \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \)
Choose a test point, e.g., \( x = \pi \):

\[ f'(\pi) = \cos(\pi) + \sin(\pi) = -1 + 0 = -1 < 0 \]

Thus, \( f(x) \) is **decreasing** in \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \).

#### Check the sign in \( (\frac{7\pi}{4}, 2\pi) \)
Choose a test point, e.g., \( x = \frac{3\pi}{2} \):

\[ f'(\frac{3\pi}{2}) = \cos(\frac{3\pi}{2}) + \sin(\frac{3\pi}{2}) = 0 - 1 = -1 < 0 \]

Thus, \( f(x) \) is **decreasing** in \( (\frac{7\pi}{4}, 2\pi) \).

### Step 5: Summarize the intervals
From our analysis, we conclude:

- The function \( f(x) \) is **strictly increasing** on the interval \( (0, \frac{3\pi}{4}) \).
- The function \( f(x) \) is **strictly decreasing** on the intervals \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \) and \( (\frac{7\pi}{4}, 2\pi) \).

### Final Answer
- **Increasing**: \( (0, \frac{3\pi}{4}) \)
- **Decreasing**: \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \) and \( (\frac{7\pi}{4}, 2\pi) \)