Find the range of `f(x)=sqrt(log(cos(sinx)))`

Answer

To find the range of the function \( f(x) = \sqrt{\log(\cos(\sin x))} \), we will follow these steps:

### Step 1: Understand the domain of the function
Since \( f(x) \) involves a square root, the expression inside the square root must be non-negative:
\[ \log(\cos(\sin x)) \geq 0 \]

### Step 2: Analyze the logarithm
The logarithm is non-negative when its argument is greater than or equal to 1:
\[ \cos(\sin x) \geq 1 \]

### Step 3: Consider the properties of the cosine function
The cosine function, \( \cos(\theta) \), has a maximum value of 1. Therefore, the only way for \( \cos(\sin x) \) to be greater than or equal to 1 is if:
\[ \cos(\sin x) = 1 \]

### Step 4: Determine when \( \cos(\sin x) = 1 \)
The cosine function equals 1 at integer multiples of \( 2\pi \):
\[ \sin x = 0 \quad \Rightarrow \quad x = n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 5: Evaluate \( f(x) \) at these points
When \( x = n\pi \):
\[ \sin(n\pi) = 0 \quad \Rightarrow \quad \cos(\sin(n\pi)) = \cos(0) = 1 \]
Thus,
\[ \log(\cos(\sin(n\pi))) = \log(1) = 0 \]
Therefore,
\[ f(n\pi) = \sqrt{0} = 0 \]

### Step 6: Conclusion about the range
Since \( f(x) \) only takes the value 0 at the points \( x = n\pi \), the range of the function \( f(x) \) is:
\[ \text{Range of } f(x) = \{0\} \]

### Final Answer
The range of \( f(x) = \sqrt{\log(\cos(\sin x))} \) is \( \{0\} \).

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