Find the magnitude of angle A, if :
`2sinAcosA-cosA-2sinA+1=0`

Answer

To solve the equation \( 2\sin A \cos A - \cos A - 2\sin A + 1 = 0 \), we will follow these steps:

### Step 1: Rearrange the equation
We start with the given equation:
\[ 2\sin A \cos A - \cos A - 2\sin A + 1 = 0 \]
We can rearrange this equation to group similar terms:
\[ 2\sin A \cos A - 2\sin A - \cos A + 1 = 0 \]

### Step 2: Factor out common terms
Notice that we can factor out \( \cos A \) from the first two terms and \( -1 \) from the last two terms:
\[ \cos A (2\sin A - 1) - (2\sin A - 1) = 0 \]
This can be rewritten as:
\[ (2\sin A - 1)(\cos A - 1) = 0 \]

### Step 3: Set each factor to zero
Now we set each factor to zero:
1. \( 2\sin A - 1 = 0 \)
2. \( \cos A - 1 = 0 \)

### Step 4: Solve for \( A \) from the first factor
From the first equation:
\[ 2\sin A = 1 \implies \sin A = \frac{1}{2} \]
The angles for which \( \sin A = \frac{1}{2} \) are:
\[ A = 30^\circ \text{ or } A = 150^\circ \]

### Step 5: Solve for \( A \) from the second factor
From the second equation:
\[ \cos A = 1 \]
The angle for which \( \cos A = 1 \) is:
\[ A = 0^\circ \]

### Step 6: Conclusion
The possible values of angle \( A \) are:
\[ A = 30^\circ, 150^\circ, \text{ and } 0^\circ \]
Since the question asks for the magnitude of angle \( A \), we can conclude that the magnitudes are:
\[ |A| = 30^\circ, 150^\circ, 0^\circ \]

### Final Answer
The magnitudes of angle \( A \) are \( 30^\circ \), \( 150^\circ \), and \( 0^\circ \).

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