Question

Solve the following equation:

\[\sin x + \sin 2x + \sin 3x + \sin 4x = 0\]

Answer 1

\[\sin x + \sin 2x + \sin 3x + \sin 4x = 0\]

\[\Rightarrow \sin3x + \sin x + \sin4x + \sin2x = 0\]
\[ \Rightarrow 2 \sin \left( \frac{4x}{2} \right) \cos \left( \frac{2x}{2} \right) + 2 \sin \left( \frac{6x}{2} \right) \cos \left( \frac{2x}{2} \right) = 0\]
\[ \Rightarrow 2 \sin2x \cos x + 2 \sin3x \cos x = 0\]
\[ \Rightarrow 2 \cos x ( \sin2x + \sin3x ) = 0\]
\[ \Rightarrow 2 \cos x\left( 2 \sin \left( \frac{5x}{2} \right) \cos \left( \frac{x}{2} \right) \right) = 0\]
\[ \Rightarrow 4 \cos x \sin \left( \frac{5x}{2} \right) \cos \left( \frac{x}{2} \right) = 0\]

\[\Rightarrow \cos x = 0 , \sin \left( \frac{5x}{2} \right) = 0\]

\[\cos \left( \frac{x}{2} \right) = 0\]

\[\Rightarrow \cos x = \cos \frac{\pi}{2}, \sin \left( \frac{5x}{2} \right) = \sin 0\] or

\[\cos \left( \frac{x}{2} \right) = \cos \frac{\pi}{2}\]

\[\Rightarrow x = (2n + 1) \frac{\pi}{2}, n \in Z or \frac{5x}{2} = n\pi , n \in Z\] or,

\[\frac{x}{2} = (2n + 1) \frac{\pi}{2} , n \in Z\]

\[\Rightarrow x = (2n + 1) \frac{\pi}{2} , n \in Z\] or

\[x = \frac{2n\pi}{5} , n \in Z\] or

\[x = (2n + 1)\pi, n \in Z\]