Calculate the value of A, if :
`cos3A.(2sin2A-1)=0`

Answer

To solve the equation \( \cos 3A (2 \sin 2A - 1) = 0 \), we can break it down into two separate cases, since a product is equal to zero if at least one of the factors is zero.

### Step 1: Set the first factor to zero
We start with the first factor:
\[ \cos 3A = 0 \]

### Step 2: Solve for \( A \)
The cosine function equals zero at odd multiples of \( 90^\circ \):
\[ 3A = 90^\circ + n \cdot 180^\circ \quad (n \in \mathbb{Z}) \]
Dividing by 3 gives:
\[ A = 30^\circ + n \cdot 60^\circ \]

### Step 3: Set the second factor to zero
Next, we consider the second factor:
\[ 2 \sin 2A - 1 = 0 \]

### Step 4: Solve for \( \sin 2A \)
Rearranging gives:
\[ 2 \sin 2A = 1 \implies \sin 2A = \frac{1}{2} \]

### Step 5: Solve for \( 2A \)
The sine function equals \( \frac{1}{2} \) at:
\[ 2A = 30^\circ + k \cdot 360^\circ \quad (k \in \mathbb{Z}) \quad \text{and} \quad 2A = 150^\circ + k \cdot 360^\circ \]
Dividing by 2 gives:
\[ A = 15^\circ + k \cdot 180^\circ \quad \text{and} \quad A = 75^\circ + k \cdot 180^\circ \]

### Step 6: Combine the solutions
From the first factor, we have:
\[ A = 30^\circ + n \cdot 60^\circ \]
From the second factor, we have:
\[ A = 15^\circ + k \cdot 180^\circ \quad \text{and} \quad A = 75^\circ + k \cdot 180^\circ \]

### Step 7: List possible values of \( A \)
Thus, the possible values of \( A \) are:
1. \( A = 30^\circ \)
2. \( A = 15^\circ \)
3. \( A = 75^\circ \)

### Final Answer
The values of \( A \) are \( 15^\circ, 30^\circ, \) and \( 75^\circ \).

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