Find the magnitude of angle A, if :
`2tan3Acos3A-tan3A+1=2cos3A`

Answer

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

### Step 1: Rearrange the equation
Start with the original equation:
\[ 2\tan(3A)\cos(3A) - \tan(3A) + 1 = 2\cos(3A) \]
Rearranging gives us:
\[ 2\tan(3A)\cos(3A) - \tan(3A) - 2\cos(3A) + 1 = 0 \]

### Step 2: Factor out common terms
Notice that we can group the terms involving \(\tan(3A)\):
\[ \tan(3A)(2\cos(3A) - 1) + (1 - 2\cos(3A)) = 0 \]
This can be rewritten as:
\[ (2\cos(3A) - 1)(\tan(3A) - 1) = 0 \]

### Step 3: Set each factor to zero
Now we have two factors that can be set to zero:
1. \( 2\cos(3A) - 1 = 0 \)
2. \( \tan(3A) - 1 = 0 \)

### Step 4: Solve the first equation
From \( 2\cos(3A) - 1 = 0 \):
\[ 2\cos(3A) = 1 \implies \cos(3A) = \frac{1}{2} \]
The angles for which \(\cos(3A) = \frac{1}{2}\) are:
\[ 3A = 60^\circ + 360^\circ n \quad \text{or} \quad 3A = 300^\circ + 360^\circ n \]
where \(n\) is any integer.

### Step 5: Solve for A
Dividing by 3 gives:
\[ A = 20^\circ + 120^\circ n \quad \text{or} \quad A = 100^\circ + 120^\circ n \]

### Step 6: Solve the second equation
From \( \tan(3A) - 1 = 0 \):
\[ \tan(3A) = 1 \]
The angles for which \(\tan(3A) = 1\) are:
\[ 3A = 45^\circ + 180^\circ n \]
Dividing by 3 gives:
\[ A = 15^\circ + 60^\circ n \]

### Step 7: Find possible values of A
From the solutions:
1. From \(\cos(3A) = \frac{1}{2}\):
- \(A = 20^\circ\) or \(A = 100^\circ\)
2. From \(\tan(3A) = 1\):
- \(A = 15^\circ\)

### Final Answer
Thus, the possible values for the angle \(A\) are:
\[ A = 15^\circ, 20^\circ, 100^\circ \]

### Magnitude of angle A
The magnitude of angle \(A\) can be stated as:
\[ \text{Magnitude of angle A} = 15^\circ, 20^\circ \]