Solve for x :
`sin^(2)x+sin^(2)30^(@)=1`

Answer

To solve the equation \( \sin^2 x + \sin^2 30^\circ = 1 \), we can follow these steps:

### Step 1: Substitute the value of \( \sin 30^\circ \)
We know that \( \sin 30^\circ = \frac{1}{2} \). Therefore, we can substitute this value into the equation.

\[ \sin^2 x + \left(\frac{1}{2}\right)^2 = 1 \]

### Step 2: Simplify the equation
Calculating \( \left(\frac{1}{2}\right)^2 \):

\[ \sin^2 x + \frac{1}{4} = 1 \]

### Step 3: Isolate \( \sin^2 x \)
Now, we will isolate \( \sin^2 x \) by subtracting \( \frac{1}{4} \) from both sides:

\[ \sin^2 x = 1 - \frac{1}{4} \]

### Step 4: Calculate the right-hand side
Calculating \( 1 - \frac{1}{4} \):

\[ \sin^2 x = \frac{4}{4} - \frac{1}{4} = \frac{3}{4} \]

### Step 5: Take the square root
Now we take the square root of both sides:

\[ \sin x = \sqrt{\frac{3}{4}} \quad \text{or} \quad \sin x = -\sqrt{\frac{3}{4}} \]

This simplifies to:

\[ \sin x = \frac{\sqrt{3}}{2} \quad \text{or} \quad \sin x = -\frac{\sqrt{3}}{2} \]

### Step 6: Find the angles
We know that \( \sin x = \frac{\sqrt{3}}{2} \) corresponds to:

\[ x = 60^\circ + 360^\circ n \quad \text{or} \quad x = 120^\circ + 360^\circ n \quad \text{for any integer } n \]

And \( \sin x = -\frac{\sqrt{3}}{2} \) corresponds to:

\[ x = 240^\circ + 360^\circ n \quad \text{or} \quad x = 300^\circ + 360^\circ n \quad \text{for any integer } n \]

### Final Answer
Thus, the solutions for \( x \) are:

\[ x = 60^\circ + 360^\circ n, \quad x = 120^\circ + 360^\circ n, \quad x = 240^\circ + 360^\circ n, \quad x = 300^\circ + 360^\circ n \quad \text{for any integer } n \]