The domain of the function `f(x)=(tan 2x)/(6 cos x+2 sin 2x)" is "`

Options

A. `R-{(2n+1)pi/2: n in z}`
B. `R-{(2n+1)pi/4: n in z}`
C. `R-{{(2n+1)pi/2: n in Z} cup {(2n+1) pi/4: n in z}}`
D. None of these

Answer

To find the domain of the function \( f(x) = \frac{\tan(2x)}{6 \cos x + 2 \sin(2x)} \), we need to ensure that the denominator does not equal zero and that the function is defined.

### Step 1: Identify the components of the function
The function is given as:
\[ f(x) = \frac{\tan(2x)}{6 \cos x + 2 \sin(2x)} \]
We know that \(\tan(2x)\) is defined for all \(x\) except where \(\cos(2x) = 0\).

### Step 2: Find where \(\cos(2x) = 0\)
The cosine function is zero at odd multiples of \(\frac{\pi}{2}\):
\[ 2x = (2n + 1) \frac{\pi}{2} \quad \text{for } n \in \mathbb{Z} \]
Dividing by 2 gives:
\[ x = \frac{(2n + 1) \pi}{4} \]

### Step 3: Find where the denominator is zero
Next, we need to ensure that the denominator \(6 \cos x + 2 \sin(2x) \neq 0\). We can rewrite \(\sin(2x)\) as \(2 \sin x \cos x\):
\[ 6 \cos x + 2(2 \sin x \cos x) = 6 \cos x + 4 \sin x \cos x \]
Factoring out \(2 \cos x\):
\[ 2 \cos x (3 + 2 \sin x) \neq 0 \]
This gives us two conditions:
1. \(2 \cos x \neq 0\) which implies \(\cos x \neq 0\)
2. \(3 + 2 \sin x \neq 0\) which implies \(\sin x \neq -\frac{3}{2}\)

### Step 4: Solve \(\cos x = 0\)
The cosine function is zero at:
\[ x = (2n + 1) \frac{\pi}{2} \quad \text{for } n \in \mathbb{Z} \]

### Step 5: Solve \(3 + 2 \sin x = 0\)
From \(3 + 2 \sin x = 0\):
\[ \sin x = -\frac{3}{2} \]
This is not possible since the sine function only takes values in the range \([-1, 1]\).

### Conclusion: Combine the results
The domain of \(f(x)\) excludes the points where \(\cos(2x) = 0\) and \(\cos x = 0\):
- Exclude \(x = \frac{(2n + 1) \pi}{4}\)
- Exclude \(x = (2n + 1) \frac{\pi}{2}\)

Thus, the domain of the function \(f(x)\) is:
\[ \mathbb{R} \setminus \left\{ \frac{(2n + 1) \pi}{4}, (2n + 1) \frac{\pi}{2} \right\} \quad \text{for } n \in \mathbb{Z} \]