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 start with the function:
\[ f(x) = \sin^2 x + 2 \sin x + 2 \]
We can rearrange this by recognizing that \( \sin^2 x + 2 \sin x \) can be expressed as a perfect square.

### Step 2: Complete the square
We can rewrite \( \sin^2 x + 2 \sin x \) as:
\[ \sin^2 x + 2 \sin x = (\sin x + 1)^2 - 1 \]
Thus, we have:
\[ f(x) = (\sin x + 1)^2 - 1 + 2 = (\sin x + 1)^2 + 1 \]

### Step 3: Determine the range of \( \sin x \)
The sine function, \( \sin x \), varies between -1 and 1:
\[ -1 \leq \sin x \leq 1 \]

### Step 4: Transform the sine range
Adding 1 to the sine function:
\[ 0 \leq \sin x + 1 \leq 2 \]

### Step 5: Square the transformed range
Now we square the entire inequality:
\[ 0 \leq (\sin x + 1)^2 \leq 4 \]

### Step 6: Shift the range
Finally, we add 1 to the squared values:
\[ 1 \leq (\sin x + 1)^2 + 1 \leq 4 + 1 \]
This simplifies to:
\[ 1 \leq f(x) \leq 5 \]

### Conclusion
Thus, the range of the function \( f(x) = \sin^2 x + 2 \sin x + 2 \) is:
\[ [1, 5] \]