The sum of all values of `theta in (0,pi/2)` satisfying `sin^(2)2theta+cos^(4)2theta=3/4` is

Options

A. `pi`
B. `(pi)/(2)`
C. `(3pi)/(8)`
D. `(5pi)/(4)`

Answer

To solve the equation \( \sin^2(2\theta) + \cos^4(2\theta) = \frac{3}{4} \) for \( \theta \) in the interval \( (0, \frac{\pi}{2}) \), we will follow these steps:

### Step 1: Rewrite the equation
We know that \( \sin^2(2\theta) + \cos^2(2\theta) = 1 \). Therefore, we can express \( \sin^2(2\theta) \) as:
\[ \sin^2(2\theta) = 1 - \cos^2(2\theta) \]
Substituting this into the original equation gives:
\[ 1 - \cos^2(2\theta) + \cos^4(2\theta) = \frac{3}{4} \]

### Step 2: Simplify the equation
Rearranging the equation, we get:
\[ \cos^4(2\theta) - \cos^2(2\theta) + 1 - \frac{3}{4} = 0 \]
This simplifies to:
\[ \cos^4(2\theta) - \cos^2(2\theta) + \frac{1}{4} = 0 \]

### Step 3: Let \( x = \cos^2(2\theta) \)
Now, let \( x = \cos^2(2\theta) \). The equation becomes:
\[ x^2 - x + \frac{1}{4} = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot \frac{1}{4}}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{1 \pm \sqrt{1 - 1}}{2} = \frac{1 \pm 0}{2} = \frac{1}{2} \]
Thus, we have:
\[ \cos^2(2\theta) = \frac{1}{2} \]

### Step 5: Solve for \( 2\theta \)
Taking the square root gives:
\[ \cos(2\theta) = \frac{1}{\sqrt{2}} \quad \text{or} \quad \cos(2\theta) = -\frac{1}{\sqrt{2}} \]
From \( \cos(2\theta) = \frac{1}{\sqrt{2}} \), we have:
\[ 2\theta = \frac{\pi}{4} + 2n\pi \quad \text{or} \quad 2\theta = -\frac{\pi}{4} + 2n\pi \]
For \( n = 0 \):
\[ 2\theta = \frac{\pi}{4} \Rightarrow \theta = \frac{\pi}{8} \]
\[ 2\theta = -\frac{\pi}{4} \Rightarrow \theta = -\frac{\pi}{8} \quad \text{(not in the interval)} \]

### Step 6: Check for other solutions
From \( \cos(2\theta) = -\frac{1}{\sqrt{2}} \):
\[ 2\theta = \frac{3\pi}{4} + 2n\pi \quad \text{or} \quad 2\theta = -\frac{3\pi}{4} + 2n\pi \]
For \( n = 0 \):
\[ 2\theta = \frac{3\pi}{4} \Rightarrow \theta = \frac{3\pi}{8} \]
\[ 2\theta = -\frac{3\pi}{4} \Rightarrow \theta = -\frac{3\pi}{8} \quad \text{(not in the interval)} \]

### Step 7: Sum the solutions
The valid solutions in the interval \( (0, \frac{\pi}{2}) \) are \( \theta = \frac{\pi}{8} \) and \( \theta = \frac{3\pi}{8} \). Therefore, the sum of all values of \( \theta \) is:
\[ \frac{\pi}{8} + \frac{3\pi}{8} = \frac{4\pi}{8} = \frac{\pi}{2} \]

### Final Answer
The sum of all values of \( \theta \) in \( (0, \frac{\pi}{2}) \) satisfying the equation is \( \frac{\pi}{2} \).

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