Question

If \(2 \tan^2 \theta - 4 \sec \theta + 3 = 0\), then \(2 \sec \theta\) is:

• A. 3
• B. \(2 + \sqrt{2}\) and \(2 - \sqrt{2}\)
• C. \(2 - \sqrt{2}\)
• D. \(2 + \sqrt{2}\)

Hint:

Always check the domain of the trigonometric function when solving equations to ensure the solutions are valid within the context given.

Answer 1

The Correct Option is D

To solve the equation \(2 \tan^2 \theta - 4 \sec \theta + 3 = 0\), substitute \(\sec \theta\) with \(\frac{1}{\cos \theta}\) and \(\tan^2 \theta\) with \(\sec^2 \theta - 1\) to obtain: \[ 2(\sec^2 \theta - 1) - 4 \sec \theta + 3 = 0 \] Simplifying, we have: \[ 2 \sec^2 \theta - 4 \sec \theta + 1 = 0 \] Let \(x = \sec \theta\). The equation becomes: \[ 2x^2 - 4x + 1 = 0 \] Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -4\), and \(c = 1\): \[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 2 \times 1}}{2 \times 2} \] \[ x = \frac{4 \pm \sqrt{16 - 8}}{4} \] \[ x = \frac{4 \pm \sqrt{8}}{4} \] \[ x = \frac{4 \pm 2\sqrt{2}}{4} \] \[ x = 1 \pm \frac{\sqrt{2}}{2} \] Thus, \(x = 1 + \frac{\sqrt{2}}{2}\) or \(x = 1 - \frac{\sqrt{2}}{2}\), translating to \(\sec \theta = 1 + \frac{\sqrt{2}}{2}\) or \(\sec \theta = 1 - \frac{\sqrt{2}}{2}\). For \(2 \sec \theta\), we get: \[ 2 \sec \theta = 2 + \sqrt{2} \quad \text{or} \quad 2 \sec \theta = 2 - \sqrt{2} \] Since \(\theta\) is an acute angle, we choose the positive value: \[ 2 \sec \theta = 2 + \sqrt{2} \]