Question

Solve the following equation:

\[\sin 2x - \sin 4x + \sin 6x = 0\]

Answer 1

\[\sin 2x - \sin 4x + \sin 6x = 0\].
\[\Rightarrow 2 \sin \left( \frac{8x}{2} \right) \cos \left( \frac{4x}{2} \right) - \sin4x = 0\]
\[ \Rightarrow 2 \sin4x \cos2x - \sin4x = 0\]
\[ \Rightarrow \sin4x ( 2 \cos2x - 1) = 0\]

\[\Rightarrow \sin 4x = 0\] or

\[2 \cos2x - 1 = 0\]

\[\Rightarrow 4x = n\pi\],

\[n \in Z\] or

\[\cos2x = \frac{1}{2} \Rightarrow \cos2x = \cos \frac{\pi}{3}\]

\[\Rightarrow x = \frac{n\pi}{4}, n \in Z\] or

\[\Rightarrow x = \frac{n\pi}{4}, n \in Z\]