Find the domain of `f(x) = (sqrt((1-sinx)))/((log)_5(1-4x^2))+cos^(-1)(1-{x})dot`

Explanation

To find the domain of the function

\[ f(x) = \frac{\sqrt{1 - \sin x}}{\log_5(1 - 4x^2)} + \cos^{-1}(1 - \{x\}) \]

we will analyze each component of the function step by step.

### Step 1: Analyze \(\sqrt{1 - \sin x}\)

The expression inside the square root must be non-negative:

\[ 1 - \sin x \geq 0 \]

This implies:

\[ \sin x \leq 1 \]

Since the sine function oscillates between -1 and 1 for all real numbers, this condition is satisfied for all \(x\). Therefore, the domain from this part is:

\[ x \in \mathbb{R} \]

### Step 2: Analyze \(\log_5(1 - 4x^2)\)

The logarithm is defined only for positive arguments, so we need:

\[ 1 - 4x^2 > 0 \]

This simplifies to:

\[ 4x^2 < 1 \implies x^2 < \frac{1}{4} \implies -\frac{1}{2} < x < \frac{1}{2} \]

### Step 3: Ensure the denominator is not zero

We also need to ensure that the logarithm does not equal zero:

\[ \log_5(1 - 4x^2) \neq 0 \]

This occurs when:

\[ 1 - 4x^2 \neq 1 \implies 4x^2 \neq 0 \implies x \neq 0 \]

### Step 4: Analyze \(\cos^{-1}(1 - \{x\})\)

The expression \(1 - \{x\}\) must be within the range of the \(\cos^{-1}\) function, which is \([-1, 1]\). Thus, we have:

\[ -1 \leq 1 - \{x\} \leq 1 \]

This leads to:

\[ 0 \leq \{x\} \leq 2 \]

The fractional part \(\{x\}\) is always in the range \([0, 1)\), so this condition is also satisfied for all \(x\).

### Step 5: Combine the conditions

Now, we will combine the conditions derived from the analysis:

1. From \(\sqrt{1 - \sin x}\): \(x \in \mathbb{R}\)
2. From \(\log_5(1 - 4x^2)\): \(-\frac{1}{2} < x < \frac{1}{2}\)
3. From the denominator condition: \(x \neq 0\)

Thus, the domain of \(f(x)\) is:

\[ \text{Domain of } f(x) = \left(-\frac{1}{2}, 0\right) \cup \left(0, \frac{1}{2}\right) \]

### Final Answer:

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

\[ \boxed{\left(-\frac{1}{2}, 0\right) \cup \left(0, \frac{1}{2}\right)} \]