Find the range of the following functions :
`y = 1/(2 sin3x cos3x)`

Answer

To find the range of the function \( y = \frac{1}{2 \sin(3x) \cos(3x)} \), we can follow these steps:

### Step 1: Simplify the Function
We start with the function:
\[ y = \frac{1}{2 \sin(3x) \cos(3x)} \]
Using the trigonometric identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \), we can rewrite the denominator:
\[ y = \frac{1}{\sin(6x)} \]

### Step 2: Identify the Range of the Sine Function
The sine function, \( \sin(6x) \), oscillates between -1 and 1. Therefore, its range is:
\[ -1 \leq \sin(6x) \leq 1 \]

### Step 3: Analyze the Reciprocal Function
Since \( y = \frac{1}{\sin(6x)} \), we need to consider the values of \( y \) based on the values of \( \sin(6x) \):
- When \( \sin(6x) = 1 \), \( y = \frac{1}{1} = 1 \).
- When \( \sin(6x) = -1 \), \( y = \frac{1}{-1} = -1 \).
- As \( \sin(6x) \) approaches 0 (but never equals 0), \( y \) approaches \( \pm \infty \).

### Step 4: Determine the Range of \( y \)
From the analysis:
- When \( \sin(6x) \) is positive (between 0 and 1), \( y \) will take values greater than 1.
- When \( \sin(6x) \) is negative (between -1 and 0), \( y \) will take values less than -1.

Thus, the range of \( y \) can be expressed as:
\[ (-\infty, -1) \cup (1, \infty) \]

### Final Answer
The range of the function \( y = \frac{1}{2 \sin(3x) \cos(3x)} \) is:
\[ (-\infty, -1) \cup (1, \infty) \]

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