Find the period of the functions (where [ * ] denotes greatest integer function) `f(x) = (sin 12 x )/(2cos^(2) 6x-1)`

Answer

To find the period of the function \( f(x) = \frac{\sin(12x)}{2\cos^2(6x) - 1} \), we can break down the problem into steps.

### Step 1: Identify the components of the function
The function consists of two parts:
1. The numerator: \( \sin(12x) \)
2. The denominator: \( 2\cos^2(6x) - 1 \)

### Step 2: Find the period of the numerator
The period of \( \sin(kx) \) is given by \( \frac{2\pi}{k} \). Here, \( k = 12 \):
\[ \text{Period of } \sin(12x) = \frac{2\pi}{12} = \frac{\pi}{6} \]

### Step 3: Find the period of the denominator
The expression \( 2\cos^2(6x) - 1 \) can be rewritten using the double angle formula for cosine:
\[ 2\cos^2(6x) - 1 = \cos(12x) \]
The period of \( \cos(kx) \) is also \( \frac{2\pi}{k} \). Here, \( k = 12 \):
\[ \text{Period of } \cos(12x) = \frac{2\pi}{12} = \frac{\pi}{6} \]

### Step 4: Determine the overall period of the function
Both the numerator and the denominator have the same period of \( \frac{\pi}{6} \). Therefore, the overall period of the function \( f(x) \) is also \( \frac{\pi}{6} \).

### Conclusion
Thus, the period of the function \( f(x) = \frac{\sin(12x)}{2\cos^2(6x) - 1} \) is:
\[ \text{Period} = \frac{\pi}{6} \]

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