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

Answer

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

### Step 1: Calculate \( \cos^2(30^\circ) \)
We know that:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \cos^2(30^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

### Step 2: Substitute \( \cos^2(30^\circ) \) into the equation
Now, substitute \( \cos^2(30^\circ) \) into the original equation:
\[ \frac{3}{4} + \sin^2(2x) = 1 \]

### Step 3: 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 of both sides
Taking the square root gives us:
\[ \sin(2x) = \frac{1}{2} \quad \text{or} \quad \sin(2x) = -\frac{1}{2} \]

### Step 5: Solve for \( 2x \)
1. For \( \sin(2x) = \frac{1}{2} \):
\[ 2x = 30^\circ + n \cdot 360^\circ \quad \text{or} \quad 2x = 150^\circ + n \cdot 360^\circ \]
where \( n \) is any integer.

2. For \( \sin(2x) = -\frac{1}{2} \):
\[ 2x = 210^\circ + n \cdot 360^\circ \quad \text{or} \quad 2x = 330^\circ + n \cdot 360^\circ \]

### Step 6: Solve for \( x \)
Now, divide all equations by 2:
1. From \( 2x = 30^\circ \):
\[ x = 15^\circ \]
2. From \( 2x = 150^\circ \):
\[ x = 75^\circ \]
3. From \( 2x = 210^\circ \):
\[ x = 105^\circ \]
4. From \( 2x = 330^\circ \):
\[ x = 165^\circ \]

### Final Solutions
Thus, the possible values of \( x \) are:
\[ x = 15^\circ, 75^\circ, 105^\circ, 165^\circ \]