Find the magnitude of angle A, if :
`sin^(2)2x+sin^(2)60^(@)=1`

Answer

To solve the equation \( \sin^2(2x) + \sin^2(60^\circ) = 1 \), we will follow these steps:

### Step 1: Substitute the value of \( \sin(60^\circ) \)
We know that:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Thus, we can calculate \( \sin^2(60^\circ) \):
\[ \sin^2(60^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

### Step 2: Rewrite the equation
Now, substitute \( \sin^2(60^\circ) \) into the original equation:
\[ \sin^2(2x) + \frac{3}{4} = 1 \]

### Step 3: Isolate \( \sin^2(2x) \)
To isolate \( \sin^2(2x) \), subtract \( \frac{3}{4} \) from both sides:
\[ \sin^2(2x) = 1 - \frac{3}{4} \]
\[ \sin^2(2x) = \frac{1}{4} \]

### Step 4: Take the square root
Now, take the square root of both sides:
\[ \sin(2x) = \pm \frac{1}{2} \]

### Step 5: Find the angles for \( \sin(2x) = \frac{1}{2} \)
The angles for which \( \sin(\theta) = \frac{1}{2} \) are:
\[ 2x = 30^\circ + n \cdot 360^\circ \quad \text{or} \quad 2x = 150^\circ + n \cdot 360^\circ \quad (n \in \mathbb{Z}) \]

### Step 6: Solve for \( x \)
Dividing by 2, we get:
1. \( x = 15^\circ + n \cdot 180^\circ \)
2. \( x = 75^\circ + n \cdot 180^\circ \)

### Step 7: Consider the principal value
For the principal value, we can take \( n = 0 \):
- From the first equation: \( x = 15^\circ \)
- From the second equation: \( x = 75^\circ \)

### Conclusion
The possible values for \( x \) are \( 15^\circ \) and \( 75^\circ \).