Find the range of the following functions:
`f(x)=[ln(sin^(-1)sqrt(x^(2)+x+1))]` [.] denotes the greatest integer function.

Answer

To find the range of the function \( f(x) = \lfloor \ln(\sin^{-1}(\sqrt{x^2 + x + 1})) \rfloor \), we will follow these steps:

### Step 1: Analyze the expression inside the logarithm
The function involves \( \sin^{-1}(\sqrt{x^2 + x + 1}) \). First, we need to determine the range of \( \sqrt{x^2 + x + 1} \).

### Step 2: Find the minimum value of \( x^2 + x + 1 \)
The expression \( x^2 + x + 1 \) is a quadratic function. To find its minimum value, we can complete the square or use the vertex formula. The vertex \( x \) of a quadratic \( ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \).

Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \):
\[ x = -\frac{1}{2 \cdot 1} = -\frac{1}{2} \]

Now substituting \( x = -\frac{1}{2} \) into the quadratic:
\[ x^2 + x + 1 = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \]

### Step 3: Determine the range of \( \sqrt{x^2 + x + 1} \)
Since \( x^2 + x + 1 \) has a minimum value of \( \frac{3}{4} \) and tends to infinity as \( x \) goes to \( \pm \infty \), the range of \( \sqrt{x^2 + x + 1} \) is:
\[ \left[\sqrt{\frac{3}{4}}, \infty\right) = \left[\frac{\sqrt{3}}{2}, \infty\right) \]

### Step 4: Find the range of \( \sin^{-1}(\sqrt{x^2 + x + 1}) \)
The function \( \sin^{-1}(y) \) is defined for \( y \) in the range \( [-1, 1] \). Since \( \sqrt{x^2 + x + 1} \) is always greater than or equal to \( \frac{\sqrt{3}}{2} \), we need to check if \( \frac{\sqrt{3}}{2} \) is within the domain of \( \sin^{-1} \).

Calculating \( \sin^{-1}(\frac{\sqrt{3}}{2}) \):
\[ \sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3} \]

As \( \sqrt{x^2 + x + 1} \) approaches infinity, \( \sin^{-1}(y) \) approaches \( \frac{\pi}{2} \).

Thus, the range of \( \sin^{-1}(\sqrt{x^2 + x + 1}) \) is:
\[ \left[\frac{\pi}{3}, \frac{\pi}{2}\right) \]

### Step 5: Find the range of \( \ln(\sin^{-1}(\sqrt{x^2 + x + 1})) \)
Now we take the natural logarithm of the range:
- The minimum value is \( \ln\left(\frac{\pi}{3}\right) \).
- The maximum value approaches \( \ln\left(\frac{\pi}{2}\right) \) as \( x \) approaches infinity.

### Step 6: Apply the greatest integer function
The final step is to apply the greatest integer function \( \lfloor \cdot \rfloor \) to the range \( \left[\ln\left(\frac{\pi}{3}\right), \ln\left(\frac{\pi}{2}\right)\right) \).

Calculating the approximate values:
- \( \frac{\pi}{3} \approx 1.047 \) so \( \ln\left(\frac{\pi}{3}\right) \approx \ln(1.047) \approx 0.046 \)
- \( \frac{\pi}{2} \approx 1.570 \) so \( \ln\left(\frac{\pi}{2}\right) \approx \ln(1.570) \approx 0.451 \)

Thus, the range of \( f(x) \) is:
\[ \lfloor \ln\left(\frac{\pi}{3}\right) \rfloor \text{ to } \lfloor \ln\left(\frac{\pi}{2}\right) \rfloor = 0 \]

### Final Answer
The range of \( f(x) \) is \( \{0\} \).