If `f: RvecS ,` defined by `f(x)=sinx-sqrt(3)cosx+1,i son to,` then find the set `Sdot`

Explanation

To find the set \( S \) for the function \( f: \mathbb{R} \to S \) defined by

\[ f(x) = \sin x - \sqrt{3} \cos x + 1, \]

we will analyze the function step by step.

### Step 1: Rewrite the function
We start by rewriting the function in a more manageable form. We can express the function as:

\[ f(x) = \sin x - \sqrt{3} \cos x + 1. \]

### Step 2: Combine sine and cosine terms
Next, we can combine the sine and cosine terms using the sine addition formula. We know that:

\[ a \sin x + b \cos x = R \sin(x + \phi), \]

where \( R = \sqrt{a^2 + b^2} \) and \( \tan \phi = \frac{b}{a} \). Here, \( a = 1 \) and \( b = -\sqrt{3} \).

Calculating \( R \):

\[ R = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2. \]

Now, we need to find \( \phi \):

\[ \tan \phi = \frac{-\sqrt{3}}{1} = -\sqrt{3}. \]

This implies \( \phi = -\frac{\pi}{3} \) (since \( \tan(-\frac{\pi}{3}) = -\sqrt{3} \)).

Thus, we can rewrite \( f(x) \):

\[ f(x) = 2 \sin\left(x - \frac{\pi}{3}\right) + 1. \]

### Step 3: Determine the range of \( f(x) \)
The sine function \( \sin\left(x - \frac{\pi}{3}\right) \) oscillates between -1 and 1. Therefore, we can find the range of \( f(x) \):

\[ \text{Minimum value of } f(x) = 2(-1) + 1 = -2 + 1 = -1, \]
\[ \text{Maximum value of } f(x) = 2(1) + 1 = 2 + 1 = 3. \]

### Step 4: Conclusion about the set \( S \)
From the calculations above, we see that the function \( f(x) \) takes values in the interval \([-1, 3]\). Thus, the set \( S \) is:

\[ S = [-1, 3]. \]

### Final Answer
The set \( S \) is \([-1, 3]\).

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