For finding range we sometimes use the quadratic equation for example `f(x)=x^(2)+2x+2` will have range `[1,oo)` as, `f(x)=x^(2)+2x+2=(x+1)^(2)+1 ge 1`
Range of `sin^(2)x+2sinx+2` is

Options

A. `[1,5]`
B. `[1,oo)`
C. `[-1,1]`
D. `(0,oo)`

Answer

To find the range of the function \( f(x) = \sin^2 x + 2 \sin x + 2 \), we can follow these steps:

### Step 1: Rewrite the function
We can rewrite the function in a form that makes it easier to analyze. Notice that:

\[ f(x) = \sin^2 x + 2 \sin x + 2 \]

This can be rearranged as:

\[ f(x) = (\sin x)^2 + 2(\sin x) + 2 \]

### Step 2: Complete the square
Next, we can complete the square for the quadratic expression in terms of \( \sin x \):

\[ f(x) = (\sin x)^2 + 2(\sin x) + 1 + 1 = (\sin x + 1)^2 + 1 \]

### Step 3: Determine the range of \( \sin x \)
The sine function, \( \sin x \), has a range of \([-1, 1]\). Therefore, we can find the range of \( \sin x + 1 \):

\[ \text{Range of } \sin x + 1 = [-1 + 1, 1 + 1] = [0, 2] \]

### Step 4: Find the range of \( (\sin x + 1)^2 \)
Now we need to find the range of \( (\sin x + 1)^2 \). Since \( \sin x + 1 \) ranges from 0 to 2, squaring this gives us:

\[ \text{Range of } (\sin x + 1)^2 = [0^2, 2^2] = [0, 4] \]

### Step 5: Add 1 to the range
Finally, we add 1 to the entire range we found in the previous step:

\[ \text{Range of } f(x) = [0 + 1, 4 + 1] = [1, 5] \]

### Conclusion
Thus, the range of the function \( f(x) = \sin^2 x + 2 \sin x + 2 \) is:

\[ \text{Range} = [1, 5] \]