`"cot" theta = "sin" 2 theta, theta ne n pi, n in Z, "if" theta` equals

Options

A. `45^(@) "or" 90^(@)`
B. `45^(@) "or" 60^(@)`
C. `90^(@) "only"`
D. `45^(@) "only"`

Answer

To solve the equation \( \cot \theta = \sin 2\theta \) under the condition \( \theta \neq n\pi \) where \( n \in \mathbb{Z} \), we will follow these steps:

### Step 1: Rewrite the equation
We know that:
\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \]
and
\[ \sin 2\theta = 2 \sin \theta \cos \theta. \]
Thus, we can rewrite the equation as:
\[ \frac{\cos \theta}{\sin \theta} = 2 \sin \theta \cos \theta. \]

### Step 2: Multiply both sides by \(\sin \theta\)
To eliminate the fraction, we multiply both sides by \(\sin \theta\) (noting that \(\sin \theta \neq 0\)):
\[ \cos \theta = 2 \sin^2 \theta \cos \theta. \]

### Step 3: Rearrange the equation
We can rearrange the equation to isolate terms:
\[ \cos \theta - 2 \sin^2 \theta \cos \theta = 0. \]
Factoring out \(\cos \theta\):
\[ \cos \theta (1 - 2 \sin^2 \theta) = 0. \]

### Step 4: Set each factor to zero
This gives us two cases to consider:
1. \( \cos \theta = 0 \)
2. \( 1 - 2 \sin^2 \theta = 0 \)

### Step 5: Solve \( \cos \theta = 0 \)
The solutions for \( \cos \theta = 0 \) are:
\[ \theta = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}. \]
However, since \( \theta \neq n\pi \), the valid solution from this case is:
\[ \theta = \frac{\pi}{2}. \]

### Step 6: Solve \( 1 - 2 \sin^2 \theta = 0 \)
Rearranging gives:
\[ 2 \sin^2 \theta = 1 \implies \sin^2 \theta = \frac{1}{2} \implies \sin \theta = \pm \frac{1}{\sqrt{2}}. \]
Thus, the solutions for \(\theta\) are:
\[ \theta = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z}. \]

### Step 7: Collect all valid solutions
From both cases, the valid solutions are:
1. \( \theta = \frac{\pi}{2} \)
2. \( \theta = \frac{\pi}{4} \)

### Final Answer
Thus, the values of \( \theta \) that satisfy the equation are:
\[ \theta = \frac{\pi}{4} \quad \text{and} \quad \theta = \frac{\pi}{2}. \]