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63. Check My Understanding What are solutions of \( \sin \theta \sin 2 \theta=0 \) on the interval \( [0,2 \pi) \) ? Select 4 choice(s) \( \pi \) \( \frac{3 \pi}{2} \) \( \frac{3 \pi}{4} \) \( \frac{\pi}{3} \) \( \frac{\pi}{4} \) \( \frac{2 \pi}{3} \) 0 \( \frac{\pi}{2} \)

Ask by John Simpson. in the United Arab Emirates
Feb 26,2025

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Tutor-Verified Answer

Answer

The solutions are \( \pi \), \( \frac{3\pi}{2} \), \( 0 \), and \( \frac{\pi}{2} \).

Solution

To solve the equation sin θ · sin 2θ = 0 on the interval [0, 2π), we use the zero product property and set each factor equal to zero: 1. sin θ = 0 For θ in [0, 2π), the solutions are: θ = 0 and θ = π 2. sin 2θ = 0 Set 2θ = kπ, where k is an integer. This gives: θ = kπ/2 For θ in [0, 2π), the valid k values are 0, 1, 2, and 3, corresponding to: θ = 0, θ = π/2, θ = π, and θ = 3π/2 Thus, the distinct solutions in [0, 2π) are: 0, π/2, π, and 3π/2 Now, based on the provided choices, the correct selections are: • π • 3π/2 • 0 • π/2

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Beyond the Answer

To solve the equation \( \sin \theta \sin 2 \theta=0 \), we can look at each factor separately. The first factor, \( \sin \theta = 0 \), gives solutions at \( \theta = 0, \pi, 2\pi \). Since we're restricted to the interval \( [0, 2\pi) \), we keep \( 0 \) and \( \pi \). For the second factor, \( \sin 2\theta = 0 \), we find solutions at \( 2\theta = n\pi \) for \( n \in \mathbb{Z} \), which results in \( \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \). Putting it all together, the solutions are \( 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \). The correct choices from your list of provided solutions are \( \pi \), \( \frac{3\pi}{2} \), \( 0 \), and \( \frac{\pi}{2} \).

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