The function `f(x)=cos^(-1)((2[|sinx|+|cosx|])/(sin^2x+2sinx+11/4))` is defined if x belongs to (where [.] represents the greatest integer function)

Options

A. `[0,(7pi)/(6)]`
B. `[0,(pi)/(6)]`
C. `[(11pi)/(6)]`
D. `[pi,2pi]`

Explanation

To determine the domain of the function \( f(x) = \cos^{-1}\left(\frac{2[\lvert \sin x \rvert + \lvert \cos x \rvert]}{\sin^2 x + 2\sin x + \frac{11}{4}}\right) \), we need to ensure that the expression inside the inverse cosine function is valid. This means that it must lie within the range \([-1, 1]\).

### Step-by-Step Solution:

**Step 1: Define the function \( g(x) \)**
Let \( g(x) = \lvert \sin x \rvert + \lvert \cos x \rvert \).

**Hint 1:** Consider how the absolute values affect the sine and cosine functions over different intervals.

**Step 2: Find the maximum and minimum values of \( g(x) \)**
The maximum value of \( g(x) \) occurs when both \( \lvert \sin x \rvert \) and \( \lvert \cos x \rvert \) are maximized. The maximum value of \( g(x) \) is \( \sqrt{2} \) (when \( x = \frac{\pi}{4} \)). The minimum value occurs when either sine or cosine is zero, giving a minimum of \( 1 \).

**Hint 2:** Use the unit circle to visualize where sine and cosine reach their maximum and minimum values.

**Step 3: Analyze the denominator**
The denominator is \( \sin^2 x + 2\sin x + \frac{11}{4} \). We need this to be positive to avoid division by zero.

**Step 4: Set the inequality**
To ensure the expression is defined, we need:
\[ \sin^2 x + 2\sin x + \frac{11}{4} > 0 \]

**Hint 3:** Consider the discriminant of the quadratic equation to determine when it is positive.

**Step 5: Rewrite the quadratic**
The quadratic can be rewritten as:
\[ \sin^2 x + 2\sin x + \frac{11}{4} = \left(\sin x + 1\right)^2 + \frac{7}{4} \]
This expression is always positive since it is a sum of squares.

**Step 6: Determine the range of \( g(x) \)**
Since \( g(x) \) can take values from \( 1 \) to \( \sqrt{2} \), we can find the greatest integer function:
\[ [\lvert \sin x \rvert + \lvert \cos x \rvert] = 1 \]

**Hint 4:** Recall that the greatest integer function takes the largest integer less than or equal to the value.

**Step 7: Set up the inequality for the inverse cosine**
Now we need:
\[ \frac{2 \cdot 1}{\sin^2 x + 2\sin x + \frac{11}{4}} \leq 1 \]

**Step 8: Solve the inequality**
This leads to:
\[ 2 \leq \sin^2 x + 2\sin x + \frac{11}{4} \]
Rearranging gives:
\[ \sin^2 x + 2\sin x + \frac{3}{4} \geq 0 \]

**Step 9: Factor the quadratic**
This can be factored or analyzed using the quadratic formula. The roots can be found, and we determine the intervals where the quadratic is non-negative.

**Step 10: Find the intervals**
The roots of the quadratic are:
\[ \sin x = -\frac{3}{2} \quad (\text{not valid}) \quad \text{and} \quad \sin x = -\frac{1}{2} \]
Thus, we find that \( \sin x \geq -\frac{1}{2} \).

**Step 11: Determine the corresponding \( x \) values**
This gives:
\[ x \in \left[0, \frac{7\pi}{6}\right] \cup \left[\frac{11\pi}{6}, 2\pi\right] \]

### Final Answer:
The function \( f(x) \) is defined for:
\[ x \in \left(0, \frac{7\pi}{6}\right) \cup \left(\frac{11\pi}{6}, 2\pi\right) \]