The ratio of the greatest value of `2-cos x+s in^2x` to its least value is `7/4` (2) `9/4` (3) `(13)/4` (4) `5/4`

Answer

To solve the problem, we need to find the ratio of the greatest value to the least value of the function \( f(x) = 2 - \cos x + \sin^2 x \).

### Step 1: Define the function
Let:
\[ f(x) = 2 - \cos x + \sin^2 x \]

### Step 2: Differentiate the function
To find the critical points, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2 - \cos x + \sin^2 x) \]
Using the derivatives of \( \cos x \) and \( \sin^2 x \):
\[ f'(x) = \sin x + 2 \sin x \cos x = \sin x (1 + 2 \cos x) \]

### Step 3: Set the derivative to zero
To find the critical points, set \( f'(x) = 0 \):
\[ \sin x (1 + 2 \cos x) = 0 \]
This gives us two cases:
1. \( \sin x = 0 \)
2. \( 1 + 2 \cos x = 0 \)

### Step 4: Solve for critical points
**Case 1:** \( \sin x = 0 \)
This occurs at:
\[ x = n\pi \quad (n \in \mathbb{Z}) \]
The simplest values are \( x = 0 \).

**Case 2:** \( 1 + 2 \cos x = 0 \)
This gives:
\[ \cos x = -\frac{1}{2} \]
This occurs at:
\[ x = \frac{2\pi}{3} + 2n\pi \quad \text{and} \quad x = \frac{4\pi}{3} + 2n\pi \]
The simplest value is \( x = \frac{2\pi}{3} \).

### Step 5: Determine the nature of critical points
We need to check the second derivative to determine whether these points are maxima or minima.

### Step 6: Find the second derivative
Differentiate \( f'(x) \):
\[ f''(x) = \cos x + 2(\cos^2 x - \sin^2 x) = \cos x + 2\cos 2x \]

### Step 7: Evaluate the second derivative at critical points
1. For \( x = 0 \):
\[ f''(0) = \cos(0) + 2\cos(0) = 1 + 2 = 3 \quad (\text{local minimum}) \]

2. For \( x = \frac{2\pi}{3} \):
\[ f''\left(\frac{2\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) + 2\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} + 2\left(-\frac{1}{2}\right) = -\frac{1}{2} - 1 = -\frac{3}{2} \quad (\text{local maximum}) \]

### Step 8: Calculate the greatest and least values
1. **Greatest value at \( x = \frac{2\pi}{3} \)**:
\[ f\left(\frac{2\pi}{3}\right) = 2 - \cos\left(\frac{2\pi}{3}\right) + \sin^2\left(\frac{2\pi}{3}\right) \]
\[ = 2 - \left(-\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)^2 = 2 + \frac{1}{2} + \frac{3}{4} = \frac{4}{2} + \frac{1}{2} + \frac{3}{4} = \frac{8}{4} + \frac{1}{2} = \frac{11}{4} + \frac{3}{4} = \frac{13}{4} \]

2. **Least value at \( x = 0 \)**:
\[ f(0) = 2 - \cos(0) + \sin^2(0) = 2 - 1 + 0 = 1 \]

### Step 9: Calculate the ratio
Now we find the ratio of the greatest value to the least value:
\[ \text{Ratio} = \frac{\text{Greatest value}}{\text{Least value}} = \frac{\frac{13}{4}}{1} = \frac{13}{4} \]

### Final Answer
The ratio of the greatest value to the least value is:
\[ \frac{13}{4} \]