Pregunta
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9. Solve the following on the interval \( x \in[0,2 \pi] \). Give exact solution if possible. 6 marks) \( \sec ^{2} x-\sec x-2=0 \)

Ask by Flynn Dunn. in Canada
Jan 24,2025

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The solutions in \([0, 2π]\) are \( \frac{\pi}{3} \), \( \pi \), and \( \frac{5\pi}{3} \).

Solución

To solve the equation \( \sec^2 x - \sec x - 2 = 0 \) for \( x \) in the interval \( [0, 2\pi] \), follow these steps: 1. **Substitute and Simplify:** Let \( y = \sec x \). The equation becomes: \[ y^2 - y - 2 = 0 \] 2. **Solve the Quadratic Equation:** Factor or use the quadratic formula to solve for \( y \): \[ y^2 - y - 2 = 0 \implies y = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} \] This gives two solutions: \[ y = 2 \quad \text{or} \quad y = -1 \] 3. **Find \( x \) from \( y \):** - **Case 1:** \( \sec x = 2 \) \[ \sec x = 2 \implies \cos x = \frac{1}{2} \] The solutions within \( [0, 2\pi] \) are: \[ x = \frac{\pi}{3}, \quad \frac{5\pi}{3} \] - **Case 2:** \( \sec x = -1 \) \[ \sec x = -1 \implies \cos x = -1 \] The solution within \( [0, 2\pi] \) is: \[ x = \pi \] 4. **Combine All Solutions:** The exact solutions in the interval \( [0, 2\pi] \) are: \[ x = \frac{\pi}{3}, \quad \pi, \quad \frac{5\pi}{3} \] **Final Answer:** The solutions in \([0, 2π]\) are ⅓ π, π, and 5⁄3 π.

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Bonus Knowledge

To solve the equation \( \sec ^{2} x - \sec x - 2 = 0 \), we can make a substitution by letting \( y = \sec x \). This transforms our equation into a quadratic form: \( y^2 - y - 2 = 0 \). Factoring gives us \( (y-2)(y+1) = 0 \), leading to solutions \( y = 2 \) and \( y = -1 \). Since \( y = \sec x \) and \( \sec x \) is defined as \( \frac{1}{\cos x} \), we only consider \( y = 2 \) because the secant function is positive in the intervals we're looking into. For \( \sec x = 2 \), we find that \( \cos x = \frac{1}{2} \), leading us to solutions \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \) within the interval \( [0, 2\pi] \). The solution set is therefore \( x = \frac{\pi}{3}, \frac{5\pi}{3} \). In 1733, mathematician Leonhard Euler explored trigonometric functions and their inverses, paving the way for further studies in calculus. His work helped clarify the relationships between angles, ratios, and the values of trigonometric functions, allowing us to solve equations like the one you just tackled easily! In real life, understanding trigonometric functions can be incredibly valuable, especially in fields like engineering, physics, and even computer graphics! For example, engineers use these principles to calculate forces and angles in structures. Plus, if you're into designing video games, knowing how to manipulate angles and shapes is key to creating immersive 3D environments!

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