Question

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

Hint:

Use sum-to-product identities to simplify trigonometric equations before solving them.

Answer 1

We are given the equation: \[ \sin \theta + \sin 3\theta + \sin 5\theta = 0 \] Step 1: Use sum-to-product identities. Recall the identity: \[ \sin x + \sin y = 2 \sin\left(\frac{x + y}{2}\right) \cos\left(\frac{x - y}{2}\right) \] We apply this identity to \( \sin \theta + \sin 5\theta \): \[ \sin \theta + \sin 5\theta = 2 \sin\left( \frac{6\theta}{2} \right) \cos\left( \frac{4\theta}{2} \right) = 2 \sin{3\theta} \cos{2\theta} \] So, the original equation becomes: \[ 2 \sin{3\theta} \cos{2\theta} + \sin{3\theta} = 0 \] Step 2: Factor out \( \sin{3\theta} \): \[ \sin{3\theta} (2 \cos{2\theta} + 1) = 0 \] Step 3: Solve for \( \sin{3\theta} = 0 \) and \( 2 \cos{2\theta} + 1 = 0 \): \[ \sin{3\theta} = 0 \quad \Rightarrow \quad 3\theta = n\pi \quad \Rightarrow \quad \theta = \frac{n\pi}{3} \] \[ 2 \cos{2\theta} + 1 = 0 \quad \Rightarrow \quad \cos{2\theta} = -\frac{1}{2} \quad \Rightarrow \quad 2\theta = 2n\pi \pm \frac{2\pi}{3} \quad \Rightarrow \quad \theta = n\pi \pm \frac{\pi}{3} \] Thus, the general solution is: \[ \theta = \frac{n\pi}{3}, \quad \theta = n\pi \pm \frac{\pi}{3} \]