If `(2 cos 2A-1) (tan 3A-1)= 0`, find all possible values of A

Answer

To solve the equation \((2 \cos 2A - 1)(\tan 3A - 1) = 0\), we need to consider two separate cases:

### Step 1: Solve \(2 \cos 2A - 1 = 0\)

1. Set the equation to zero:
\[ 2 \cos 2A - 1 = 0 \]
2. Rearranging gives:
\[ 2 \cos 2A = 1 \]
3. Dividing both sides by 2:
\[ \cos 2A = \frac{1}{2} \]

### Step 2: Find the angles for \(\cos 2A = \frac{1}{2}\)

1. The angles for which \(\cos \theta = \frac{1}{2}\) are:
\[ 2A = \frac{\pi}{3} + 2n\pi \quad \text{and} \quad 2A = \frac{5\pi}{3} + 2n\pi \]
where \(n\) is any integer.
2. Solving for \(A\):
\[ A = \frac{\pi}{6} + n\pi \quad \text{and} \quad A = \frac{5\pi}{6} + n\pi \]

### Step 3: Solve \(\tan 3A - 1 = 0\)

1. Set the equation to zero:
\[ \tan 3A - 1 = 0 \]
2. Rearranging gives:
\[ \tan 3A = 1 \]

### Step 4: Find the angles for \(\tan 3A = 1\)

1. The angles for which \(\tan \theta = 1\) are:
\[ 3A = \frac{\pi}{4} + n\pi \]
where \(n\) is any integer.
2. Solving for \(A\):
\[ A = \frac{\pi}{12} + \frac{n\pi}{3} \]

### Step 5: Combine the results

The possible values of \(A\) from both cases are:
1. From \(2 \cos 2A - 1 = 0\):
\[ A = \frac{\pi}{6} + n\pi \quad \text{and} \quad A = \frac{5\pi}{6} + n\pi \]
2. From \(\tan 3A - 1 = 0\):
\[ A = \frac{\pi}{12} + \frac{n\pi}{3} \]

### Final Answer

The complete set of possible values for \(A\) is:
\[ A = \frac{\pi}{6} + n\pi, \quad A = \frac{5\pi}{6} + n\pi, \quad A = \frac{\pi}{12} + \frac{n\pi}{3} \]
where \(n\) belongs to the set of integers.

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