Find the magnitude of angle A, if :
`2cos^(2)A-3cosA+1=0`

Answer

To solve the equation \(2\cos^2 A - 3\cos A + 1 = 0\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2\cos^2 A - 3\cos A + 1 = 0 \]

### Step 2: Factor the quadratic equation
We will factor the quadratic expression. We can rewrite it as:
\[ 2\cos^2 A - 2\cos A - \cos A + 1 = 0 \]
Now, we can group the terms:
\[ (2\cos A - 1)(\cos A - 1) = 0 \]

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

### Step 4: Solve for \(\cos A\)
From the first equation:
\[ 2\cos A = 1 \implies \cos A = \frac{1}{2} \]
From the second equation:
\[ \cos A = 1 \]

### Step 5: Find angles corresponding to \(\cos A\)
Now we find the angles \(A\):
1. For \(\cos A = \frac{1}{2}\):
- The angles are \(A = 60^\circ\) and \(A = 300^\circ\) (in the range of 0° to 360°).

2. For \(\cos A = 1\):
- The angle is \(A = 0^\circ\).

### Step 6: Compile the results
Thus, the possible values for angle \(A\) are:
- \(A = 0^\circ\)
- \(A = 60^\circ\)
- \(A = 300^\circ\)

### Step 7: State the final answer
The magnitude of angle \(A\) can be:
- \(0^\circ\)
- \(60^\circ\)
- \(300^\circ\)