The interval of increase of the function `y= x-2 sin x " if " 0 le x le 2pi`, is

Options

A. `(0, pi)`
B. `(0, pi/2)`
C. `(pi//2, pi)`
D. `(pi//3, 5pi//3)`

Answer

To find the interval of increase for the function \( y = x - 2 \sin x \) on the interval \( 0 \leq x \leq 2\pi \), we need to follow these steps:

### Step 1: Differentiate the function
We start by finding the derivative of the function \( y \).

\[ \frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(2 \sin x) \]

Using the differentiation rules:
- The derivative of \( x \) is \( 1 \).
- The derivative of \( \sin x \) is \( \cos x \), so the derivative of \( 2 \sin x \) is \( 2 \cos x \).

Thus, we have:

\[ \frac{dy}{dx} = 1 - 2 \cos x \]

### Step 2: Set the derivative greater than zero
To find the intervals where the function is increasing, we set the derivative greater than zero:

\[ 1 - 2 \cos x > 0 \]

### Step 3: Solve the inequality
Rearranging the inequality gives:

\[ 1 > 2 \cos x \]

Dividing both sides by 2:

\[ \frac{1}{2} > \cos x \]

This can also be written as:

\[ \cos x < \frac{1}{2} \]

### Step 4: Determine the angles where \( \cos x < \frac{1}{2} \)
The cosine function is less than \( \frac{1}{2} \) in the intervals:

\[ \left(\frac{\pi}{3}, \frac{5\pi}{3}\right) \]

### Step 5: Check the interval within \( 0 \leq x \leq 2\pi \)
Now we check the interval \( \left(\frac{\pi}{3}, \frac{5\pi}{3}\right) \) within the given range \( 0 \leq x \leq 2\pi \).

Thus, the interval of increase for the function \( y = x - 2 \sin x \) on \( 0 \leq x \leq 2\pi \) is:

\[ \boxed{\left(\frac{\pi}{3}, \frac{5\pi}{3}\right)} \]