X = 0, pi/2, 3pi/2, 2pi

Given the function values and their corresponding \( x \) values:

• At \( x = 0 \), \( F(x) = 1 \)
• At \( x = \frac{\pi}{2} \), \( F(x) = 0 \)
• At \( x = \frac{3\pi}{2} \), \( F(x) = -1 \)
• At \( x = 2\pi \), \( F(x) = 0 \)

These values suggest a trigonometric function, specifically a sine function, because it oscillates through these values over intervals of \( \frac{\pi}{2} \). To write a sine function that fits this data, we can identify the following:

1. Amplitude: The function oscillates between -1 and 1, so the amplitude \( A = 1 \).

2. Period: The values repeat every \( 2\pi \) (from \( 0 \) to \( 2\pi \)), so the period \( P = 2\pi \). The standard sine function has a period of \( 2\pi \).

3. Phase Shift (c): Since \( F(0) = 1 \), which is the maximum, we recognize that the sine function usually starts at zero. This indicates a phase shift might be present. We can arrange the standard sine function as follows: \[ F(x) = A \sin(B(x - C)) + D \] where:

   • \( A \) is the amplitude,
   • \( B \) affects the period,
   • \( C \) is the phase shift,
   • \( D \) is the vertical shift.

Given that the sine function normally starts at zero, we can shift the cosine function instead, which starts at its maximum. This leads us to consider: \[ F(x) = A \cos(Bx - C) + D \]

4. Determine \( D \): Since the function oscillates between -1 and 1, there is no vertical shift, so \( D = 0 \).

Thus, we start with: \[ F(x) = A \cos(Bx - C) \]

5. Determine the parameters:

• Since \( F(x) \) has an amplitude of 1, \( A = 1 \).
• The period remains \( P = 2\pi \). For a cosine function, \( P = \frac{2\pi}{B} \), so: \[ B = 1 \]
• The phase shift \( C \) is determined so that \( F(0) = 1 \). Given the cosine function starts at its maximum: \[ C = 0 \]

Thus, we find: \[ F(x) = \cos(x) \]

Now, let's verify with the original function values for \( F(x) = \cos(x) \):

• \( F(0) = \cos(0) = 1 \)
• \( F\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \)
• \( F\left(\frac{3\pi}{2}\right) = \cos\left(\frac{3\pi}{2}\right) = -1 \)
• \( F(2\pi) = \cos(2\pi) = 1 \)

All the results match the given function values.

Therefore, the function is: \[ F(x) = \cos(x) \] with the phase shift \( c = 0 \).