The domain of definition of the function `f(x)=sin^(-1)((4)/(3+2 cos x))`, is

Options

A. `[2npi-(pi)/(3),2 npi+(pi)/(3)], n in Z`
B. `[0,2npi+(pi)/(6)], n in Z `
C. `[2npi-(pi)/(6),0], n in Z`
D. `(2n pi-(pi)/(6), 2n pi+(pi)/(6)), n in Z `

Answer

To find the domain of the function \( f(x) = \sin^{-1}\left(\frac{4}{3 + 2 \cos x}\right) \), we need to ensure that the argument of the inverse sine function is valid. The inverse sine function \( \sin^{-1}(a) \) is defined for \( -1 \leq a \leq 1 \). Therefore, we need to solve the following inequalities:

1. \( -1 \leq \frac{4}{3 + 2 \cos x} \leq 1 \)

### Step 1: Solve the first inequality \( \frac{4}{3 + 2 \cos x} \geq -1 \)

Starting with the first inequality:

\[ \frac{4}{3 + 2 \cos x} \geq -1 \]

Multiplying both sides by \( 3 + 2 \cos x \) (noting that \( 3 + 2 \cos x > 0 \) for all \( x \) since \( \cos x \) ranges from -1 to 1):

\[ 4 \geq -1(3 + 2 \cos x) \]

This simplifies to:

\[ 4 \geq -3 - 2 \cos x \]

Rearranging gives:

\[ 2 \cos x \geq -7 \quad \Rightarrow \quad \cos x \geq -\frac{7}{2} \]

Since \( \cos x \) is always between -1 and 1, this inequality is satisfied for all \( x \).

### Step 2: Solve the second inequality \( \frac{4}{3 + 2 \cos x} \leq 1 \)

Now, we solve the second inequality:

\[ \frac{4}{3 + 2 \cos x} \leq 1 \]

Again, multiplying both sides by \( 3 + 2 \cos x \):

\[ 4 \leq 3 + 2 \cos x \]

Rearranging gives:

\[ 2 \cos x \geq 1 \quad \Rightarrow \quad \cos x \geq \frac{1}{2} \]

### Step 3: Determine the values of \( x \) for \( \cos x \geq \frac{1}{2} \)

The cosine function is equal to \( \frac{1}{2} \) at:

\[ x = \frac{\pi}{3} + 2n\pi \quad \text{and} \quad x = -\frac{\pi}{3} + 2n\pi \quad (n \in \mathbb{Z}) \]

Thus, \( \cos x \geq \frac{1}{2} \) occurs in the intervals:

\[ x \in [2n\pi - \frac{\pi}{3}, 2n\pi + \frac{\pi}{3}] \quad (n \in \mathbb{Z}) \]

### Final Domain

The domain of the function \( f(x) \) is therefore:

\[ \text{Domain of } f(x) = [2n\pi - \frac{\pi}{3}, 2n\pi + \frac{\pi}{3}] \quad (n \in \mathbb{Z}) \]