Question

Solve the following equation:

\[\sin 3x - \sin x = 4 \cos^2 x - 2\]

Answer 1

\[\sin3x - \sin x = 4 \cos^2 x - 2\]

\[\Rightarrow \sin3x - \sin x = 2 ( 2 \cos^2 x - 1)\]
\[ \Rightarrow 2 \sin \left( \frac{2x}{2} \right) \cos \left( \frac{4x}{2} \right) = 2 \cos 2x\]
\[ \Rightarrow 2 \sin x \cos2x = 2 \cos2x\]
\[ \Rightarrow \sin x \cos2x = \cos2x\]
\[ \Rightarrow \cos2x ( \sin x - 1) = 0 \]

\[\Rightarrow \cos 2x = 0\] or

\[\sin x - 1 = 0\]

⇒ \[\cos 2x = \cos \frac{\pi}{2}\] or

\[\sin x = 1 \Rightarrow \sin x = \sin \frac{\pi}{2}\]

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

\[n \in Z\] or

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

\[\Rightarrow x = (2n \hspace{0.167em} + 1)\frac{\pi}{4} , n \in Z\] or

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