If ` sqrt(3) sin x - cos x ="min"_(alpha in R){2,e^(2),pi, alpha^(2)-4alpha +7}`, then

Options

A. (a)` x =2n pi, n in I `
B. (b)`x=2n pi +(2pi)/(3) , n in I `
C. (c)`x=npi +(-1)^(n)""(pi)/(4)+(pi)/(6), n in I `
D. (d)`x=npi +(-1)^(n)""(pi)/(4)-(pi)/(3), n in I `

Answer

To solve the equation \( \sqrt{3} \sin x - \cos x = \min_{\alpha \in \mathbb{R}} \{ 2, e^2, \pi, \alpha^2 - 4\alpha + 7 \} \), we will follow these steps:

### Step 1: Analyze the Right-Hand Side (RHS)

The RHS consists of four expressions: \( 2 \), \( e^2 \), \( \pi \), and \( \alpha^2 - 4\alpha + 7 \). We need to find the minimum value of these expressions.

1. **Evaluate \( e^2 \)**:
- Since \( e \approx 2.718 \), we have \( e^2 \approx 7.389 \).

2. **Evaluate \( \pi \)**:
- \( \pi \approx 3.14 \).

3. **Analyze \( \alpha^2 - 4\alpha + 7 \)**:
- This can be rewritten as \( (\alpha - 2)^2 + 3 \).
- The minimum value of \( (\alpha - 2)^2 \) is \( 0 \) (when \( \alpha = 2 \)), so the minimum value of \( \alpha^2 - 4\alpha + 7 \) is \( 3 \).

Now we can summarize the values:
- \( 2 \)
- \( 3 \) (from \( \alpha^2 - 4\alpha + 7 \))
- \( e^2 \approx 7.389 \)
- \( \pi \approx 3.14 \)

### Step 2: Determine the Minimum Value

From the evaluations:
- The minimum value among \( 2, 3, e^2, \pi \) is \( 2 \).

Thus, we have:
\[ \sqrt{3} \sin x - \cos x = 2 \]

### Step 3: Solve the Left-Hand Side (LHS)

We can rewrite the equation:
\[ \sqrt{3} \sin x - \cos x = 2 \]

To solve for \( x \), we can rearrange it:
\[ \sqrt{3} \sin x = \cos x + 2 \]

### Step 4: Normalize the Equation

Divide both sides by 2:
\[ \frac{\sqrt{3}}{2} \sin x - \frac{1}{2} \cos x = 1 \]

### Step 5: Use Trigonometric Identities

Recognize that:
\[ \frac{\sqrt{3}}{2} = \sin \frac{\pi}{3}, \quad \text{and} \quad \frac{1}{2} = \cos \frac{\pi}{3} \]

Thus, we can apply the sine subtraction formula:
\[ \sin\left(x - \frac{\pi}{3}\right) = 1 \]

### Step 6: Solve for \( x \)

The general solution for \( \sin \theta = 1 \) is:
\[ \theta = \frac{\pi}{2} + 2n\pi, \quad n \in \mathbb{Z} \]

Substituting back:
\[ x - \frac{\pi}{3} = \frac{\pi}{2} + 2n\pi \]

Solving for \( x \):
\[ x = \frac{\pi}{2} + \frac{\pi}{3} + 2n\pi = \frac{3\pi}{6} + \frac{2\pi}{6} + 2n\pi = \frac{5\pi}{6} + 2n\pi \]

### Step 7: Final Form of the Solution

Thus, the solutions for \( x \) are:
\[ x = \frac{5\pi}{6} + 2n\pi, \quad n \in \mathbb{Z} \]