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: Group the terms
We can rearrange the equation to group the terms involving \(\cos A\) and \(\sin A\):
\[ 2\sin A \cos A - \cos A - 2\sin A + 1 = 0 \]

### Step 2: Factor out common terms
We can factor out \(\cos A\) from the first two terms:
\[ \cos A (2\sin A - 1) - 2\sin A + 1 = 0 \]

### Step 3: Rearrange the equation
Now, we can rearrange the equation:
\[ \cos A (2\sin A - 1) = 2\sin A - 1 \]

### Step 4: Set the equation to zero
Now we can set the equation to zero:
\[ \cos A (2\sin A - 1) - (2\sin A - 1) = 0 \]

### Step 5: Factor the equation
We can factor the equation:
\[ (2\sin A - 1)(\cos A - 1) = 0 \]

### Step 6: Solve for \(\cos A\) and \(\sin A\)
Now we have two equations to solve:
1. \(2\sin A - 1 = 0\)
2. \(\cos A - 1 = 0\)

From the first equation:
\[ 2\sin A = 1 \implies \sin A = \frac{1}{2} \]
From the second equation:
\[ \cos A = 1 \]

### Step 7: Find the angles
Now we can find the angles corresponding to these values:
1. For \(\sin A = \frac{1}{2}\), the angle \(A\) can be \(30^\circ\) (or \(\frac{\pi}{6}\) radians).
2. For \(\cos A = 1\), the angle \(A\) is \(0^\circ\).

### Conclusion
Thus, the possible values for angle \(A\) are:
\[ A = 0^\circ \text{ or } 30^\circ \]