let `f:[-pi/3,pi/6]rarr B` defined by `f(x)=2cos^2x+sqrt3sin2x+1`. Find B such that `f^-1` exists. Also find `f^-1`

Explanation

To solve the problem, we need to find the set \( B \) such that the function \( f \) is invertible, and then we will find the inverse function \( f^{-1} \).

### Step 1: Define the function
The function is given by:
\[ f(x) = 2\cos^2 x + \sqrt{3}\sin 2x + 1 \]
where \( x \) is in the interval \( \left[-\frac{\pi}{3}, \frac{\pi}{6}\right] \).

### Step 2: Simplify the function
Using the double angle identity for sine, we can rewrite \( \sin 2x \) as \( 2\sin x \cos x \). Thus, we can express \( f(x) \) as:
\[ f(x) = 2\cos^2 x + \sqrt{3}(2\sin x \cos x) + 1 \]
This simplifies to:
\[ f(x) = 2\cos^2 x + 2\sqrt{3}\sin x \cos x + 1 \]

### Step 3: Rewrite using trigonometric identities
We can use the identity \( 2\cos^2 x = \cos 2x + 1 \) to rewrite the function:
\[ f(x) = \cos 2x + 1 + 2\sqrt{3}\sin x \cos x + 1 \]
This can be further simplified to:
\[ f(x) = \cos 2x + 2 + 2\sqrt{3}\sin x \cos x \]
Now, we can express \( 2\sqrt{3}\sin x \cos x \) as \( \sqrt{3}\sin 2x \):
\[ f(x) = \cos 2x + \sqrt{3}\sin 2x + 2 \]

### Step 4: Find the range of \( f(x) \)
To find the range of \( f(x) \), we need to analyze the expression:
\[ f(x) = 2 + \sqrt{(1^2 + (\sqrt{3})^2)}\sin(2x + \phi) \]
where \( \phi \) is the phase shift. The maximum value of \( \sin \) function is 1 and the minimum is -1. Thus:
\[ f(x) = 2 + 2\sin(2x + \phi) \]
This means the range of \( f(x) \) is:
\[ [2 - 2, 2 + 2] = [0, 4] \]

### Step 5: Set \( B \)
Since \( f \) is one-to-one and onto, the set \( B \) such that \( f^{-1} \) exists is:
\[ B = [0, 4] \]

### Step 6: Find the inverse function \( f^{-1}(x) \)
To find the inverse function, we set:
\[ y = f(x) = 2 + \sqrt{3}\sin(2x + \phi) \]
Rearranging gives:
\[ \sqrt{3}\sin(2x + \phi) = y - 2 \]
Thus:
\[ \sin(2x + \phi) = \frac{y - 2}{\sqrt{3}} \]
Taking the inverse sine:
\[ 2x + \phi = \sin^{-1}\left(\frac{y - 2}{\sqrt{3}}\right) \]
Now isolating \( x \):
\[ 2x = \sin^{-1}\left(\frac{y - 2}{\sqrt{3}}\right) - \phi \]
Finally:
\[ x = \frac{1}{2}\left(\sin^{-1}\left(\frac{y - 2}{\sqrt{3}}\right) - \phi\right) \]
Replacing \( y \) with \( x \) gives us the inverse function:
\[ f^{-1}(x) = \frac{1}{2}\left(\sin^{-1}\left(\frac{x - 2}{\sqrt{3}}\right) - \phi\right) \]

### Summary
Thus, the set \( B \) is \( [0, 4] \) and the inverse function \( f^{-1}(x) \) can be expressed as:
\[ f^{-1}(x) = \frac{1}{2}\left(\sin^{-1}\left(\frac{x - 2}{\sqrt{3}}\right) - \phi\right) \]