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

Answer

To find the magnitude of angle A from the equation \(2\sin^2 A - 3\sin A + 1 = 0\), we can follow these steps:

### Step 1: Rewrite the equation
The given equation is a quadratic in terms of \(\sin A\). We can rewrite it as:
\[ 2\sin^2 A - 3\sin A + 1 = 0 \]

### Step 2: Identify coefficients
In the quadratic equation \(ax^2 + bx + c = 0\), we identify:
- \(a = 2\)
- \(b = -3\)
- \(c = 1\)

### Step 3: Factor the quadratic
To factor the quadratic, we can look for two numbers that multiply to \(ac = 2 \cdot 1 = 2\) and add up to \(b = -3\). The numbers are \(-2\) and \(-1\). Thus, we can rewrite the equation:
\[ 2\sin^2 A - 2\sin A - \sin A + 1 = 0 \]
Now, we can factor by grouping:
\[ 2\sin A(\sin A - 1) - 1(\sin A - 1) = 0 \]
This can be factored as:
\[ (2\sin A - 1)(\sin A - 1) = 0 \]

### Step 4: Set each factor to zero
Now, we set each factor equal to zero:
1. \(2\sin A - 1 = 0\)
2. \(\sin A - 1 = 0\)

### Step 5: Solve for \(\sin A\)
From the first equation:
\[ 2\sin A = 1 \implies \sin A = \frac{1}{2} \]
From the second equation:
\[ \sin A = 1 \]

### Step 6: Find the angles
Now we find the angles corresponding to these sine values:
1. \(\sin A = \frac{1}{2}\) occurs at \(A = 30^\circ\) (and also \(A = 150^\circ\) in the range of \(0^\circ\) to \(360^\circ\)).
2. \(\sin A = 1\) occurs at \(A = 90^\circ\).

### Step 7: List all possible angles
Therefore, the possible values for angle \(A\) are:
- \(A = 30^\circ\)
- \(A = 90^\circ\)
- (and \(A = 150^\circ\) if considering angles beyond the first quadrant).

### Final Answer
The magnitudes of angle \(A\) are \(30^\circ\), \(90^\circ\), and \(150^\circ\).

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