If `cos^(-1)(n)/(2pi)gt(2pi)/(3)` then maximum and minimum values of integer in are respectively

Options

A. 3
B. 4
C. `-4`
D. `-3`

Answer

To solve the problem, we start with the given inequality:

\[ \frac{\cos^{-1}(n)}{2\pi} > \frac{2\pi}{3} \]

### Step 1: Multiply both sides by \(2\pi\)

Since \(2\pi\) is a positive number, we can multiply both sides of the inequality without changing the direction of the inequality:

\[ \cos^{-1}(n) > \frac{2\pi}{3} \cdot 2\pi \]

### Step 2: Simplify the right side

Calculating the right side:

\[ \frac{2\pi}{3} \cdot 2\pi = \frac{4\pi^2}{3} \]

So we have:

\[ \cos^{-1}(n) > \frac{4\pi^2}{3} \]

### Step 3: Apply the cosine function

To isolate \(n\), we apply the cosine function to both sides. Since the cosine function is decreasing in the range of \([0, \pi]\), we need to reverse the inequality:

\[ n < \cos\left(\frac{4\pi^2}{3}\right) \]

### Step 4: Evaluate \(\cos\left(\frac{4\pi^2}{3}\right)\)

To find \(\cos\left(\frac{4\pi^2}{3}\right)\), we need to recognize that \(\frac{4\pi^2}{3}\) is outside the standard range for cosine. We can reduce this angle:

\[ \frac{4\pi^2}{3} = 2\pi + \frac{\pi}{3} \]

Thus, we have:

\[ \cos\left(\frac{4\pi^2}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 5: Establish the bounds for \(n\)

Now we have:

\[ n < \frac{1}{2} \]

### Step 6: Determine the integer values

Since \(n\) is an integer, the maximum integer value for \(n\) that satisfies this inequality is:

\[ n_{\text{max}} = 0 \]

The minimum integer value for \(n\) can be any integer less than \(\frac{1}{2}\), which can go down to negative infinity. However, we are interested in the largest integer less than or equal to \(0\):

\[ n_{\text{min}} = -1 \]

### Conclusion

Thus, the maximum and minimum integer values of \(n\) are:

\[ \text{Maximum value of } n = 0 \]
\[ \text{Minimum value of } n = -1 \]