Let `f:[-pi/3,(2pi)/3]vec[0,4]` be a function defined as `f(x)=sqrt(3)sinx-cosx+2.` Then `f^(-1)(x)` is given by `sin^(-1)((x-2)/2)-pi/6` `sin^(-1)((x-2)/2)+pi/6` `(2pi)/3+cos^(-1)((x-2)/2)` (d) none of these

Options

A. `sin^(-1)((x-2)/(2))-(pi)/(6)`
B. `sin^(-1)((x-2)/(2))+(pi)/(6)`
C. `(2pi)/(3)+cos^(-1)((x-2)/(2))`
D. None of these

Explanation

To find the inverse of the function \( f(x) = \sqrt{3} \sin x - \cos x + 2 \) defined on the interval \( [-\frac{\pi}{3}, \frac{2\pi}{3}] \) and mapping to \( [0, 4] \), we can follow these steps:

### Step 1: Set up the equation
We start by letting \( y = f(x) \):
\[ y = \sqrt{3} \sin x - \cos x + 2 \]

### Step 2: Rearrange the equation
We can rearrange the equation to isolate the trigonometric functions:
\[ y - 2 = \sqrt{3} \sin x - \cos x \]

### Step 3: Express in terms of sine
Next, we can express the right-hand side in terms of sine. We know that:
\[ \sqrt{3} \sin x - \cos x = 2 \left( \frac{\sqrt{3}}{2} \sin x - \frac{1}{2} \cos x \right) \]
This can be rewritten using the sine of a difference:
\[ \sqrt{3} \sin x - \cos x = 2 \sin\left(x - \frac{\pi}{6}\right) \]
Thus, we have:
\[ y - 2 = 2 \sin\left(x - \frac{\pi}{6}\right) \]

### Step 4: Solve for sine
Dividing both sides by 2 gives:
\[ \frac{y - 2}{2} = \sin\left(x - \frac{\pi}{6}\right) \]

### Step 5: Take the inverse sine
Now, we take the inverse sine of both sides:
\[ x - \frac{\pi}{6} = \sin^{-1}\left(\frac{y - 2}{2}\right) \]

### Step 6: Solve for x
Finally, we solve for \( x \):
\[ x = \sin^{-1}\left(\frac{y - 2}{2}\right) + \frac{\pi}{6} \]

### Step 7: Replace y with x
Since we are finding the inverse function \( f^{-1}(x) \), we replace \( y \) with \( x \):
\[ f^{-1}(x) = \sin^{-1}\left(\frac{x - 2}{2}\right) + \frac{\pi}{6} \]

### Conclusion
Thus, the correct answer is:
\[ f^{-1}(x) = \sin^{-1}\left(\frac{x - 2}{2}\right) + \frac{\pi}{6} \]