`f(x)=log_(5)(cos^(-1)sqrt(x^(2)+5x+6)+sec^(-1)[{x}+1]([.]` denotes greatest integer function and `{.}` denotes fractional part function. Find the domain of `f(x)`:

Options

A. `[(-5-sqrt(5))/(2),-3]cup[-2,(-5+sqrt(5))/(2)]`
B. `[(-sqrt(5)+5)/(2),-3]cup[-2,(5+sqrt(5))/(2))`
C. `((-5+sqrt(5))/(4),-3]cup[-2,(-5+sqrt(5))/(4))`
D. `((-5-sqrt(5))/(2),-3]cup[-2,(-5+sqrt(5))/(2))`

Explanation

To find the domain of the function \( f(x) = \log_{5} \left( \cos^{-1} \sqrt{x^{2} + 5x + 6} + \sec^{-1} \left( \lfloor x \rfloor + 1 \right) \right) \), we need to ensure that the arguments of both the logarithm and the inverse trigonometric functions are valid.

### Step 1: Analyze the expression inside the logarithm

The function \( \log_{5}(y) \) is defined for \( y > 0 \). Therefore, we need:
\[ \cos^{-1} \sqrt{x^{2} + 5x + 6} + \sec^{-1} \left( \lfloor x \rfloor + 1 \right) > 0 \]

### Step 2: Determine the domain of \( \cos^{-1} \)

The function \( \cos^{-1}(z) \) is defined for \( z \) in the range \( [0, 1] \). Thus, we need:
\[ 0 \leq \sqrt{x^{2} + 5x + 6} \leq 1 \]

### Step 3: Solve the inequality \( \sqrt{x^{2} + 5x + 6} \leq 1 \)

Squaring both sides:
\[ x^{2} + 5x + 6 \leq 1 \]
\[ x^{2} + 5x + 5 \leq 0 \]

### Step 4: Find the roots of the quadratic equation

To find the roots of \( x^{2} + 5x + 5 = 0 \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 - 20}}{2} = \frac{-5 \pm \sqrt{5}}{2} \]

Thus, the roots are:
\[ x = \frac{-5 + \sqrt{5}}{2}, \quad x = \frac{-5 - \sqrt{5}}{2} \]

### Step 5: Analyze the intervals

The quadratic \( x^{2} + 5x + 5 \) opens upwards (since the coefficient of \( x^{2} \) is positive). The inequality \( x^{2} + 5x + 5 \leq 0 \) holds between the roots:
\[ \frac{-5 - \sqrt{5}}{2} \leq x \leq \frac{-5 + \sqrt{5}}{2} \]

### Step 6: Determine the domain of \( \sec^{-1} \)

The function \( \sec^{-1}(z) \) is defined for \( |z| \geq 1 \). Therefore, we need:
\[ |\lfloor x \rfloor + 1| \geq 1 \]

This implies:
\[ \lfloor x \rfloor + 1 \leq -1 \quad \text{or} \quad \lfloor x \rfloor + 1 \geq 1 \]
Which simplifies to:
\[ \lfloor x \rfloor \leq -2 \quad \text{or} \quad \lfloor x \rfloor \geq 0 \]

### Step 7: Combine the conditions

From the analysis, we have two conditions:
1. \( \frac{-5 - \sqrt{5}}{2} \leq x \leq \frac{-5 + \sqrt{5}}{2} \)
2. \( \lfloor x \rfloor \leq -2 \) or \( \lfloor x \rfloor \geq 0 \)

### Step 8: Find the intersection of these intervals

Calculating the approximate values:
- \( \frac{-5 - \sqrt{5}}{2} \approx -4.618 \)
- \( \frac{-5 + \sqrt{5}}{2} \approx -0.382 \)

Thus, the interval for \( x \) is approximately \( [-4.618, -0.382] \).

### Final Domain

Combining the conditions, the domain of \( f(x) \) is:
\[ \left[-5 - \frac{\sqrt{5}}{2}, -3\right] \cup \left[-2, -5 + \frac{\sqrt{5}}{2}\right] \]

### Conclusion

Thus, the domain of \( f(x) \) is:
\[ \left[-5 - \frac{\sqrt{5}}{2}, -3\right] \cup \left[-2, -5 + \frac{\sqrt{5}}{2}\right] \]