Answer
The solutions for θ are:
\[
\theta = \frac{\pi}{8} + \frac{k\pi}{2} \quad \text{or} \quad \theta = \frac{1}{2} \arctan\left(\frac{1}{2}\right) + \frac{k\pi}{2} \quad \text{for any integer } k
\]
Solution
To solve the equation:
\[ 2 \sec^{2}(2\theta) = 3 \tan(2\theta) + 1 \]
follow these steps:
### Step 1: Rewrite in Terms of \(\tan(2\theta)\)
Use the trigonometric identity:
\[ \sec^{2}(x) = 1 + \tan^{2}(x) \]
Apply this to the equation:
\[
2 \sec^{2}(2\theta) = 3 \tan(2\theta) + 1 \\
2 (1 + \tan^{2}(2\theta)) = 3 \tan(2\theta) + 1
\]
### Step 2: Simplify the Equation
Expand and rearrange the equation:
\[
2 + 2 \tan^{2}(2\theta) = 3 \tan(2\theta) + 1 \\
2 \tan^{2}(2\theta) - 3 \tan(2\theta) + 1 = 0
\]
### Step 3: Solve the Quadratic Equation
Let \( x = \tan(2\theta) \). The equation becomes:
\[
2x^{2} - 3x + 1 = 0
\]
Solve for \( x \) using the quadratic formula:
\[
x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}
\]
Thus, the solutions for \( x \) are:
\[
x = \frac{4}{4} = 1 \quad \text{and} \quad x = \frac{2}{4} = \frac{1}{2}
\]
So,
\[
\tan(2\theta) = 1 \quad \text{or} \quad \tan(2\theta) = \frac{1}{2}
\]
### Step 4: Find \( \theta \) for Each Case
**Case 1:** \( \tan(2\theta) = 1 \)
\[
2\theta = \frac{\pi}{4} + k\pi \quad \Rightarrow \quad \theta = \frac{\pi}{8} + \frac{k\pi}{2}, \quad k \in \mathbb{Z}
\]
**Case 2:** \( \tan(2\theta) = \frac{1}{2} \)
\[
2\theta = \arctan\left(\frac{1}{2}\right) + k\pi \quad \Rightarrow \quad \theta = \frac{1}{2} \arctan\left(\frac{1}{2}\right) + \frac{k\pi}{2}, \quad k \in \mathbb{Z}
\]
### Final Solution
All real numbers θ that satisfy the equation are given by:
\[
\boxed{
\theta = \frac{\pi}{8} + \frac{k\pi}{2} \quad \text{or} \quad \theta = \frac{1}{2} \arctan\left(\frac{1}{2}\right) + \frac{k\pi}{2} \quad \text{for any integer } k
}
\]
Reviewed and approved by the UpStudy tutoring team
Explain
Simplify this solution
