Question
Solve the following equations:
$\sin2\text{x}-\sin4\text{x}+\sin6\text{x}=0$
$\sin2\text{x}-\sin4\text{x}+\sin6\text{x}=0$
✓
Answer
$\sin2\text{x}-\sin4\text{x}+\sin6\text{x}=0$
$(\sin2\text{x}+\sin6\text{x})-\sin4\text{x}=0$
$2.\sin\Big(\frac{8\text{x}}{2}\Big).\cos\Big(\frac{4\text{x}}{2}\Big)-\sin4\text{x}=0$
$2\sin4\text{x}.\cos2\text{x}-\sin4\text{x}=0$
$\sin4\text{x}(2\cos2\text{x}-1)=0$
$\sin4\text{x}=0$ or $2\cos2\text{x}-1=0$
$4\text{x}=\text{n}(\pi)$ or $\cos2\text{x}=\frac{1}{2}$
$\text{x}=[\frac{\text{n}\pi}{4}]$ or $\cos2\text{x}=\cos[\frac{\pi}{3}]$
$\text{x}=[\frac{\text{n}\pi}{4}]$ or $\text{x}=\text{n}(\pi)\pm[\frac{\pi}{6}]$
$(\sin2\text{x}+\sin6\text{x})-\sin4\text{x}=0$
$2.\sin\Big(\frac{8\text{x}}{2}\Big).\cos\Big(\frac{4\text{x}}{2}\Big)-\sin4\text{x}=0$
$2\sin4\text{x}.\cos2\text{x}-\sin4\text{x}=0$
$\sin4\text{x}(2\cos2\text{x}-1)=0$
$\sin4\text{x}=0$ or $2\cos2\text{x}-1=0$
$4\text{x}=\text{n}(\pi)$ or $\cos2\text{x}=\frac{1}{2}$
$\text{x}=[\frac{\text{n}\pi}{4}]$ or $\cos2\text{x}=\cos[\frac{\pi}{3}]$
$\text{x}=[\frac{\text{n}\pi}{4}]$ or $\text{x}=\text{n}(\pi)\pm[\frac{\pi}{6}]$
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