Question

Find the general solution of \( \sin \theta + \sin 3\theta + \sin 5\theta = 0 \).

Hint:

Use trigonometric identities like sum-to-product to simplify multiple sine terms.

Answer 1

Group terms: \( \sin \theta + \sin 5\theta + \sin 3\theta = 0 \).
Use sum-to-product identity for \( \sin \theta + \sin 5\theta \): \[ \sin \theta + \sin 5\theta = 2 \sin \left( \frac{\theta + 5\theta}{2} \right) \cos \left( \frac{5\theta - \theta}{2} \right) = 2 \sin 3\theta \cos 2\theta. \] Thus: \[ 2 \sin 3\theta \cos 2\theta + \sin 3\theta = 0. \] Factor: \[ \sin 3\theta (2 \cos 2\theta + 1) = 0. \] So, either: \[ \sin 3\theta = 0 \quad \text{or} \quad 2 \cos 2\theta + 1 = 0. \] - Case 1: \( \sin 3\theta = 0 \Rightarrow 3\theta = n\pi \Rightarrow \theta = \frac{n\pi}{3} \).
- Case 2: \( 2 \cos 2\theta + 1 = 0 \Rightarrow \cos 2\theta = -\frac{1}{2} \Rightarrow 2\theta = \pi \pm \frac{\pi}{3} + 2k\pi \Rightarrow \theta = \frac{\pi}{2} \pm \frac{\pi}{6} + k\pi \).
Thus, \( \theta = \frac{n\pi}{3} \) or \( \theta = \frac{\pi}{2} \pm \frac{\pi}{6} + k\pi \).
General solution: \[ \theta = \frac{n\pi}{3}, \quad \theta = \frac{2k+1}{2}\pi \pm \frac{\pi}{6}, \quad n, k \in \mathbb{Z}. \]