Question

Solve the following equation:
\[\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x\]

Answer 1

\[\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x\]
\[ \Rightarrow 2\sin2x\cos x - 3\sin2x = 2\cos2x\cos x - 3\cos2x\]
\[ \Rightarrow \sin2x\left( 2\cos x - 3 \right) = \cos2x\left( 2\cos x - 3 \right)\]
\[ \Rightarrow \left( \sin2x - \cos2x \right)\left( 2\cos x - 3 \right) = 0\]
\[\Rightarrow \sin2x - \cos2x = 0 or 2\cos x - 3 = 0\]
\[ \Rightarrow \sin2x = \cos2x or \cos x = \frac{3}{2}\]
\[ \Rightarrow \tan2x = 1 or \cos x = \frac{3}{2}\]
But,
\[\cos x = \frac{3}{2}\] is not possible.

\[\left( - 1 \leq \cos x \leq 1 \right)\]

\[\therefore \tan2x = 1 = \tan\frac{\pi}{4}\]

\[ \Rightarrow 2x = n\pi + \frac{\pi}{4}, n \in Z\]

\[ \Rightarrow x = \frac{n\pi}{2} + \frac{\pi}{8}, n \in Z\]