Let `f(x)=(sqrt(sinx))/(1+(sinx)^(1/3))` then domain `f` contains

Options

A. `(0,pi)`
B. `(-2pi,-pi)`
C. `(3pi,4pi)`
D. `(4pi,6pi)`

Answer

To find the domain of the function \( f(x) = \frac{\sqrt{\sin x}}{1 + (\sin x)^{1/3}} \), we need to consider the conditions under which this function is defined.

### Step 1: Determine when the square root is defined
The square root function \( \sqrt{\sin x} \) is defined when \( \sin x \geq 0 \). Therefore, we have:
\[ \sin x \geq 0 \]

### Step 2: Determine when the denominator is defined
The denominator \( 1 + (\sin x)^{1/3} \) must not be equal to zero:
\[ 1 + (\sin x)^{1/3} \neq 0 \]
This simplifies to:
\[ (\sin x)^{1/3} \neq -1 \]
Cubing both sides gives:
\[ \sin x \neq -1 \]

### Step 3: Combine the conditions
From the above steps, we have two conditions:
1. \( \sin x \geq 0 \)
2. \( \sin x \neq -1 \)

### Step 4: Analyze the sine function
The sine function is non-negative in the intervals:
- \( [0, \pi] \)
- \( [2\pi, 3\pi] \)
- \( [4\pi, 5\pi] \)
- ...

The sine function equals \(-1\) at \( x = \frac{3\pi}{2} + 2k\pi \) for any integer \( k \).

### Step 5: Identify the intervals where the conditions hold
- In the interval \( [0, \pi] \), \( \sin x \) is non-negative and does not equal \(-1\).
- In the interval \( [-2\pi, -\pi] \), \( \sin x \) is also non-negative and does not equal \(-1\).
- In the interval \( [3\pi, 4\pi] \), \( \sin x \) is negative, so this interval is excluded.
- In the interval \( [4\pi, 5\pi] \), \( \sin x \) is again non-negative.

### Conclusion
Thus, the domain of \( f(x) \) contains the intervals:
- \( [0, \pi] \)
- \( [-2\pi, -\pi] \)
- \( [4\pi, 5\pi] \)

### Final Answer
The domain of \( f \) contains:
- \( [0, \pi] \)
- \( [-2\pi, -\pi] \)