The number of integers in the range of `3 sin^(2)x+3sin x cos x+7cos^(2)x` is

Options

A. 3
B. 4
C. 5
D. 6

Explanation

To find the number of integers in the range of the expression \(3 \sin^2 x + 3 \sin x \cos x + 7 \cos^2 x\), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ y = 3 \sin^2 x + 3 \sin x \cos x + 7 \cos^2 x \]
We can express \(\sin^2 x\) and \(\cos^2 x\) in terms of \(\cos 2x\):
\[ \sin^2 x = \frac{1 - \cos 2x}{2}, \quad \cos^2 x = \frac{1 + \cos 2x}{2} \]
Substituting these into the expression gives:
\[ y = 3 \left(\frac{1 - \cos 2x}{2}\right) + 3 \left(\frac{\sin 2x}{2}\right) + 7 \left(\frac{1 + \cos 2x}{2}\right) \]

### Step 2: Simplify the expression
Now we simplify:
\[ y = \frac{3(1 - \cos 2x) + 3 \sin 2x + 7(1 + \cos 2x)}{2} \]
\[ = \frac{3 - 3 \cos 2x + 3 \sin 2x + 7 + 7 \cos 2x}{2} \]
\[ = \frac{10 + 4 \cos 2x + 3 \sin 2x}{2} \]
\[ = 5 + 2 \cos 2x + \frac{3}{2} \sin 2x \]

### Step 3: Identify the range of the trigonometric part
The expression \(2 \cos 2x + \frac{3}{2} \sin 2x\) can be rewritten in the form \(R \sin(2x + \phi)\) where:
\[ R = \sqrt{(2)^2 + \left(\frac{3}{2}\right)^2} = \sqrt{4 + \frac{9}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2} \]
Thus, the maximum and minimum values of \(2 \cos 2x + \frac{3}{2} \sin 2x\) are \(\frac{5}{2}\) and \(-\frac{5}{2}\) respectively.

### Step 4: Determine the range of \(y\)
Adding 5 to the maximum and minimum values:
- Maximum: \(5 + \frac{5}{2} = 5 + 2.5 = 7.5\)
- Minimum: \(5 - \frac{5}{2} = 5 - 2.5 = 2.5\)

Thus, the range of \(y\) is:
\[ [2.5, 7.5] \]

### Step 5: Identify the integers in the range
The integers in the range \([2.5, 7.5]\) are:
- 3
- 4
- 5
- 6
- 7

### Conclusion
The total number of integers in this range is 5.