The general solution of the equation `tan2theta tan3theta=1` is

Options

A. `theta = (4n+1)pi/10, n ne 4k-2, k in I`
B. `theta = (2n+3)pi/10, n ne 5k+2, k in I`
C. `theta = (4n+3) pi/10, n ne 5k, k in l`
D. `theta = (2n+1)pi/10, n ne 5k-3, k in I`

Answer

To solve the equation \( \tan 2\theta \tan 3\theta = 1 \), we will follow these steps:

### Step 1: Rewrite the equation in terms of sine and cosine
We know that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Therefore, we can rewrite the equation as:

\[ \tan 2\theta \tan 3\theta = \frac{\sin 2\theta}{\cos 2\theta} \cdot \frac{\sin 3\theta}{\cos 3\theta} = 1 \]

### Step 2: Simplify the equation
This gives us:

\[ \frac{\sin 2\theta \sin 3\theta}{\cos 2\theta \cos 3\theta} = 1 \]

Multiplying both sides by \( \cos 2\theta \cos 3\theta \) leads to:

\[ \sin 2\theta \sin 3\theta = \cos 2\theta \cos 3\theta \]

### Step 3: Rearranging the equation
We can rearrange this equation to:

\[ \sin 2\theta \sin 3\theta - \cos 2\theta \cos 3\theta = 0 \]

### Step 4: Use the cosine of sum formula
Using the identity \( \cos(a + b) = \cos a \cos b - \sin a \sin b \), we can rewrite the equation as:

\[ -\cos(2\theta + 3\theta) = 0 \]

This simplifies to:

\[ -\cos(5\theta) = 0 \]

### Step 5: Solve for \( \theta \)
From \( -\cos(5\theta) = 0 \), we have:

\[ \cos(5\theta) = 0 \]

The general solution for \( \cos x = 0 \) is given by:

\[ x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \]

Substituting \( 5\theta \) for \( x \):

\[ 5\theta = \frac{\pi}{2} + n\pi \]

### Step 6: Isolate \( \theta \)
Dividing both sides by 5 gives:

\[ \theta = \frac{\pi}{10} + \frac{n\pi}{5} \]

### Step 7: Write the final general solution
We can express this in a more compact form:

\[ \theta = \frac{\pi}{10} + \frac{n\pi}{5} = \frac{\pi}{10} + \frac{2n\pi}{10} = \frac{(2n + 1)\pi}{10}, \quad n \in \mathbb{Z} \]

Thus, the general solution of the equation \( \tan 2\theta \tan 3\theta = 1 \) is:

\[ \theta = \frac{(2n + 1)\pi}{10}, \quad n \in \mathbb{Z} \]