Find the maximum value of `sqrt(3)sinx+cosxa n dx` for which a maximum value occurs.

Explanation

To find the maximum value of the expression \( \sqrt{3} \sin x + \cos x \), we can follow these steps:

### Step 1: Rewrite the expression
We can express \( \sqrt{3} \sin x + \cos x \) in a form that makes it easier to analyze. We can factor it out as follows:

\[ \sqrt{3} \sin x + \cos x = 2 \left( \frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x \right) \]

### Step 2: Identify coefficients
Now, we recognize that \( \frac{\sqrt{3}}{2} \) and \( \frac{1}{2} \) correspond to the sine and cosine of specific angles. Specifically, we can relate these to the sine of an angle:

\[ \frac{\sqrt{3}}{2} = \sin\left(\frac{\pi}{3}\right) \quad \text{and} \quad \frac{1}{2} = \cos\left(\frac{\pi}{3}\right) \]

### Step 3: Use the angle addition formula
Using the angle addition formula for sine, we can rewrite the expression as:

\[ \sqrt{3} \sin x + \cos x = 2 \left( \sin\left(x + \frac{\pi}{3}\right) \right) \]

### Step 4: Determine the maximum value
The sine function achieves its maximum value of 1. Therefore, the maximum value of our expression occurs when:

\[ \sin\left(x + \frac{\pi}{3}\right) = 1 \]

### Step 5: Solve for \( x \)
This occurs when:

\[ x + \frac{\pi}{3} = \frac{\pi}{2} + 2k\pi \quad \text{for integer } k \]

Solving for \( x \):

\[ x = \frac{\pi}{2} - \frac{\pi}{3} + 2k\pi = \frac{3\pi}{6} - \frac{2\pi}{6} + 2k\pi = \frac{\pi}{6} + 2k\pi \]

### Step 6: Calculate the maximum value
Substituting back into the expression for the maximum value:

\[ \text{Maximum value} = 2 \cdot 1 = 2 \]

### Final Result
Thus, the maximum value of \( \sqrt{3} \sin x + \cos x \) is \( 2 \), and it occurs at \( x = \frac{\pi}{6} + 2k\pi \).

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