b \( 2 \sec ^{2} 2 \theta=3 \tan 2 \theta+1 \)

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: \[ \boxed{ \theta = \frac{\pi}{8} + \frac{k\pi}{2} \quad \text{or} \quad \theta = \frac{1}{2} \arctan\left(\frac{1}{2}\right) + \frac{k\pi}{2} \quad \text{for any integer } k } \]