Solve $sin 2 x - sin 4 x + sin 6 x = 0$

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Solution

Step 1: Simplification
Given $sin 2 x - sin 4 x + sin 6 x = 0$
$\Rightarrow (sin 2 x + sin 6 x) - sin 4 x = 0$
$\Rightarrow 2 sin \frac{6 x + 2 x}{2} cos \frac{6 x - 2 x}{2} - sin 4 x = 0$
$\Rightarrow 2 sin \frac{8 x}{2} cos \frac{4 x}{2} - sin 4 x = 0$
$\Rightarrow 2 sin 4 x cos 2 x - sin 4 x = 0$
$\Rightarrow sin 4 x (2 cos 2 x - 1) = 0$
$\Rightarrow sin 4 x = 0, 2 cos 2 x = 1$

We need to find general solution for both separately.
Step 2: General solution of $sin 4 x = 0$
General solution is $4 x = n π \Rightarrow x = \frac{n π}{4}$ where $n \in z$

Step 3: General solution of $cos 2 x = \frac{1}{2}$
$\Rightarrow cos 2 x = cos (\frac{π}{3})$
$∴ 2 x = 2 n π \pm \frac{π}{3} \Rightarrow x = n π \pm \frac{π}{6}$ where $n \in z$

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Find the general solutions of sin 2x - sin 4x + sin 6x = 0.

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x

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