Mathematics

Mathematics

Problem

Determine the general solution of 2sin⁡xcos⁡x−sin⁡x=02\sin x\cos x - \sin x = 0. a) x=nπorx=2nπ±π6,  n∈Zx = n\pi \quad \text{or} \quad x = 2n\pi \pm \dfrac{\pi}{6},\; n\in\mathbb{Z} b) x=nπorx=2nπ±π3,  n∈Zx = n\pi \quad \text{or} \quad x = 2n\pi \pm \dfrac{\pi}{3},\; n\in\mathbb{Z} c) x=nπorx=nπ±π3,  n∈Zx = n\pi \quad \text{or} \quad x = n\pi \pm \dfrac{\pi}{3},\; n\in\mathbb{Z} d) x=nπorx=2nπ±2π3,  n∈Zx = n\pi \quad \text{or} \quad x = 2n\pi \pm \dfrac{2\pi}{3},\; n\in\mathbb{Z}
Intuitive explanation

This is a trigonometric equation where the key is to recognize that you have a common factor hiding in plain sight. Factor out sin⁡x\sin x from both terms, then use the double-angle identity 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \sin 2x to simplify what remains. Once factored, you'll have a product equal to zero—solve each factor separately to find all solutions.

Intuitive explanation

This is a trigonometric equation where the key is to recognize that you have a common factor hiding in plain sight. Factor out sin⁡x\sin x from both terms, then use the double-angle identity 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \sin 2x to simplify what remains. Once factored, you'll have a product equal to zero—solve each factor separately to find all solutions.

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