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Example 2.4.2. Given \( \sin 2 \theta=\frac{-1}{\sqrt{2}} \), solve for \( \theta \). Example 2.4.3. Given \( 2 \sin ^{2} \theta+\sin \theta-1=0 \), solve for \( \theta \).

Ask by Fernandez Stuart. in South Africa
Mar 13,2025

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**Example 2.4.2** Solutions for \( \sin 2\theta = \frac{-1}{\sqrt{2}} \) are: \[ \theta = \frac{5\pi}{8} + \pi k \quad \text{or} \quad \theta = \frac{7\pi}{8} + \pi k,\quad k\in\mathbb{Z}. \] **Example 2.4.3** Solutions for \( 2 \sin^2 \theta + \sin \theta - 1 = 0 \) are: \[ \theta = \frac{\pi}{6} + 2\pi k,\quad \theta = \frac{5\pi}{6} + 2\pi k,\quad \theta = \frac{3\pi}{2} + 2\pi k,\quad k\in\mathbb{Z}. \]

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To solve \( \sin 2\theta = \frac{-1}{\sqrt{2}} \), remember that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Thus, you can find the angles whose sine value is \( \frac{-1}{\sqrt{2}} \), which occurs at \( 2\theta = \frac{7\pi}{4} + 2k\pi \) or \( \frac{5\pi}{4} + 2k\pi \) for any integer \( k \). Divide these by 2 to find \( \theta \) in the ranges of \( 0 \) to \( 2\pi \). For the equation \( 2 \sin^2 \theta + \sin \theta - 1 = 0 \), treat it as a quadratic equation in terms of \( \sin \theta \). Using the quadratic formula \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2, b = 1, c = -1 \), yields the solutions for \( \sin \theta \). You can find the corresponding angles in the desired range of \( 0 \) to \( 2\pi \).

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