Sin Solution 2Dca4E

1. **State the problem:** Find all solutions to the equation $$\sin \theta + \sin 4\theta = \sin 2\theta + \sin 3\theta$$ in the interval $$-\pi < \theta \leq \pi$$ and determine the multiplicity of the solution $$\theta = 0$$. 2. **Use sum-to-product identities:** Recall that $$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$$. 3. **Apply to both sides:** $$\sin \theta + \sin 4\theta = 2 \sin \left( \frac{\theta + 4\theta}{2} \right) \cos \left( \frac{\theta - 4\theta}{2} \right) = 2 \sin \left( \frac{5\theta}{2} \right) \cos \left( -\frac{3\theta}{2} \right)$$ Since $$\cos(-x) = \cos x$$, this is $$2 \sin \left( \frac{5\theta}{2} \right) \cos \left( \frac{3\theta}{2} \right)$$. Similarly, $$\sin 2\theta + \sin 3\theta = 2 \sin \left( \frac{2\theta + 3\theta}{2} \right) \cos \left( \frac{2\theta - 3\theta}{2} \right) = 2 \sin \left( \frac{5\theta}{2} \right) \cos \left( -\frac{\theta}{2} \right) = 2 \sin \left( \frac{5\theta}{2} \right) \cos \left( \frac{\theta}{2} \right)$$. 4. **Rewrite the equation:** $$2 \sin \left( \frac{5\theta}{2} \right) \cos \left( \frac{3\theta}{2} \right) = 2 \sin \left( \frac{5\theta}{2} \right) \cos \left( \frac{\theta}{2} \right)$$ 5. **Divide both sides by 2:** $$\sin \left( \frac{5\theta}{2} \right) \cos \left( \frac{3\theta}{2} \right) = \sin \left( \frac{5\theta}{2} \right) \cos \left( \frac{\theta}{2} \right)$$ 6. **Bring all terms to one side:** $$\sin \left( \frac{5\theta}{2} \right) \cos \left( \frac{3\theta}{2} \right) - \sin \left( \frac{5\theta}{2} \right) \cos \left( \frac{\theta}{2} \right) = 0$$ 7. **Factor out common term:** $$\sin \left( \frac{5\theta}{2} \right) \left( \cos \left( \frac{3\theta}{2} \right) - \cos \left( \frac{\theta}{2} \right) \right) = 0$$ 8. **Set each factor equal to zero:** - Case 1: $$\sin \left( \frac{5\theta}{2} \right) = 0$$ - Case 2: $$\cos \left( \frac{3\theta}{2} \right) - \cos \left( \frac{\theta}{2} \right) = 0$$ 9. **Solve Case 1:** $$\sin \left( \frac{5\theta}{2} \right) = 0 \implies \frac{5\theta}{2} = k\pi, \quad k \in \mathbb{Z}$$ So, $$\theta = \frac{2k\pi}{5}$$ Within $$-\pi < \theta \leq \pi$$, find all integer $$k$$ such that $$-\pi < \frac{2k\pi}{5} \leq \pi$$. Divide by $$\pi$$: $$-1 < \frac{2k}{5} \leq 1 \implies -\frac{5}{2} < k \leq \frac{5}{2}$$ So $$k = -2, -1, 0, 1, 2$$. Corresponding $$\theta$$ values: $$\theta = \frac{2(-2)\pi}{5} = -\frac{4\pi}{5}, \quad -\frac{2\pi}{5}, \quad 0, \quad \frac{2\pi}{5}, \quad \frac{4\pi}{5}$$. 10. **Solve Case 2:** $$\cos \left( \frac{3\theta}{2} \right) = \cos \left( \frac{\theta}{2} \right)$$ Recall that $$\cos A = \cos B \implies A = B + 2m\pi \text{ or } A = -B + 2m\pi, m \in \mathbb{Z}$$. So, - Subcase 2a: $$\frac{3\theta}{2} = \frac{\theta}{2} + 2m\pi \implies \frac{3\theta}{2} - \frac{\theta}{2} = 2m\pi \implies \theta = 2m\pi$$ Within $$-\pi < \theta \leq \pi$$, $$\theta = 0$$ only. - Subcase 2b: $$\frac{3\theta}{2} = -\frac{\theta}{2} + 2m\pi \implies \frac{3\theta}{2} + \frac{\theta}{2} = 2m\pi \implies 2\theta = 2m\pi \implies \theta = m\pi$$ Within $$-\pi < \theta \leq \pi$$, $$\theta = -\pi, 0, \pi$$. 11. **Combine all solutions:** From Case 1: $$\theta = -\frac{4\pi}{5}, -\frac{2\pi}{5}, 0, \frac{2\pi}{5}, \frac{4\pi}{5}$$ From Case 2: $$\theta = -\pi, 0, \pi$$ Combine and remove duplicates: $$\theta = -\pi, -\frac{4\pi}{5}, -\frac{2\pi}{5}, 0, \frac{2\pi}{5}, \frac{4\pi}{5}, \pi$$ 12. **Multiplicity of $$\theta = 0$$:** To find multiplicity, check the order of zero of the function $$f(\theta) = \sin \theta + \sin 4\theta - \sin 2\theta - \sin 3\theta$$ at $$\theta = 0$$. Calculate derivatives: - $$f(0) = 0$$ - $$f'(\theta) = \cos \theta + 4 \cos 4\theta - 2 \cos 2\theta - 3 \cos 3\theta$$ Evaluate at 0: $$f'(0) = 1 + 4 - 2 - 3 = 0$$ - $$f''(\theta) = -\sin \theta - 16 \sin 4\theta + 4 \sin 2\theta + 9 \sin 3\theta$$ Evaluate at 0: $$f''(0) = 0$$ - $$f'''(\theta) = -\cos \theta - 64 \cos 4\theta + 8 \cos 2\theta + 27 \cos 3\theta$$ Evaluate at 0: $$f'''(0) = -1 - 64 + 8 + 27 = -30 \neq 0$$ Since $$f(0) = f'(0) = f''(0) = 0$$ but $$f'''(0) \neq 0$$, the root at $$\theta = 0$$ has multiplicity 3. **Final answer:** $$\boxed{\theta = -\pi, -\frac{4\pi}{5}, -\frac{2\pi}{5}, 0, \frac{2\pi}{5}, \frac{4\pi}{5}, \pi}$$ with the solution $$\theta = 0$$ having multiplicity 3.