Find the domain of
`y=sqrt(log_(3)(cos(sinx)))`

Answer

To find the domain of the function \( y = \sqrt{\log_{3}(\cos(\sin x))} \), we need to ensure that the expression inside the square root is non-negative, as the square root function is only defined for non-negative values. Additionally, we need to ensure that the logarithm is defined and positive.

### Step 1: Ensure the argument of the logarithm is positive
The logarithm \( \log_{3}(\cos(\sin x)) \) is defined only when \( \cos(\sin x) > 0 \).

### Step 2: Determine when \( \cos(\sin x) > 0 \)
The cosine function is positive in the intervals:
- \( 2k\pi < \sin x < (2k+1)\pi \) for \( k \in \mathbb{Z} \)

However, since \( \sin x \) takes values between -1 and 1, we need to find the values of \( \sin x \) that keep \( \cos(\sin x) > 0 \).

### Step 3: Find the intervals for \( \sin x \)
The cosine function is positive for:
- \( -\frac{\pi}{2} < \sin x < \frac{\pi}{2} \)

### Step 4: Solve for \( x \)
To find the values of \( x \) that satisfy \( -\frac{\pi}{2} < \sin x < \frac{\pi}{2} \):
- Since \( \sin x \) oscillates between -1 and 1, we need to find the intervals of \( x \) where \( \sin x \) is within these bounds.

### Step 5: Find the domain
The sine function is periodic with a period of \( 2\pi \). Therefore, the intervals where \( \sin x \) remains in the range \( (-1, 1) \) are:
- \( x \in \left(2k\pi - \frac{\pi}{2}, 2k\pi + \frac{\pi}{2}\right) \) for \( k \in \mathbb{Z} \)

### Step 6: Ensure the logarithm is non-negative
Next, we need to ensure \( \log_{3}(\cos(\sin x)) \geq 0 \), which means:
- \( \cos(\sin x) \geq 1 \)

This condition is satisfied when \( \sin x = 0 \), which occurs at:
- \( x = n\pi \) for \( n \in \mathbb{Z} \)

### Final Domain
Thus, the domain of the function \( y = \sqrt{\log_{3}(\cos(\sin x))} \) is:
- \( x \in \{ n\pi \mid n \in \mathbb{Z} \} \)