The function `f(x)=sqrt(3)sin x-cos x` will increase monotically in the interval(s)

Options

A. `0le x le (pi)/(2)`
B. `-(pi)/(3)le x le (2pi)/(3)`
C. `(5pi)/(3)le x le(8pi)/(3)`
D. `0 le x le pi`

Explanation

To determine the intervals in which the function \( f(x) = \sqrt{3} \sin x - \cos x \) increases monotonically, we will follow these steps:

### Step 1: Find the derivative of the function
To find where the function is increasing, we need to compute the derivative \( f'(x) \).

\[ f'(x) = \sqrt{3} \cos x + \sin x \]

### Step 2: Set the derivative greater than zero
The function \( f(x) \) is increasing where \( f'(x) > 0 \).

\[ \sqrt{3} \cos x + \sin x > 0 \]

### Step 3: Rearrange the inequality
We can rearrange the inequality:

\[ \sin x > -\sqrt{3} \cos x \]

### Step 4: Divide both sides by \( \cos x \) (where \( \cos x \neq 0 \))
This gives us:

\[ \tan x > -\sqrt{3} \]

### Step 5: Determine the angles
The inequality \( \tan x > -\sqrt{3} \) corresponds to angles where the tangent function is greater than \( -\sqrt{3} \). The reference angle for \( \tan x = -\sqrt{3} \) is \( -\frac{\pi}{3} \).

### Step 6: Identify the intervals
The tangent function is periodic with a period of \( \pi \). Thus, we can find the intervals where \( \tan x > -\sqrt{3} \):

1. The first interval starts from \( -\frac{\pi}{3} \) to \( \frac{2\pi}{3} \).
2. The second interval starts from \( \frac{2\pi}{3} + \pi \) to \( \frac{5\pi}{3} + \pi \), which is \( \frac{5\pi}{3} \) to \( \frac{8\pi}{3} \).

### Step 7: Combine the intervals
Thus, the function \( f(x) \) increases monotonically in the intervals:

\[ \left[-\frac{\pi}{3}, \frac{2\pi}{3}\right] \quad \text{and} \quad \left[\frac{5\pi}{3}, \frac{8\pi}{3}\right] \]

### Final Answer
The function \( f(x) = \sqrt{3} \sin x - \cos x \) increases monotonically in the intervals \( \left[-\frac{\pi}{3}, \frac{2\pi}{3}\right] \) and \( \left[\frac{5\pi}{3}, \frac{8\pi}{3}\right] \).

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