Question

Solve the following equation:

\[\cos x + \sin x = \cos 2x + \sin 2x\]

Answer 1

\[\cos x + \sin x = \cos2x + \sin2x\]

\[\Rightarrow \cos x - \cos2x = \sin2x - \sin x\]
\[ \Rightarrow - 2 \sin \left( \frac{3x}{2} \right) \sin \left( \frac{- x}{2} \right) = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{3x}{2} \right)\]
\[ \Rightarrow 2 \sin \left( \frac{3x}{2} \right) \sin \left( \frac{x}{2} \right) = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{3x}{2} \right)\]
\[ \Rightarrow 2 \sin \left( \frac{x}{2} \right) \left[ \sin \left( \frac{3x}{2} \right) - \cos \left( \frac{3x}{2} \right) \right] = 0\]

\[\Rightarrow \sin \frac{x}{2} = 0\] or

\[\sin \frac{3x}{2} - \cos \frac{3x}{2} = 0\] 

\[\Rightarrow \sin \frac{x}{2} = \sin 0\] or

\[\sin \frac{3x}{2} = \cos \frac{3x}{2}\]

⇒ \[\frac{x}{2} = n\pi\],

\[n \in Z\] or

\[\cos \frac{3x}{2} = \cos \left( \frac{\pi}{2} - \frac{3x}{2} \right)\]

\[\Rightarrow x = 2n\pi, n \in Z\] or

\[\frac{3x}{2} = 2m\pi \pm \left( \frac{\pi}{2} - \frac{3x}{2} \right), m \in Z\]

⇒ \[x = 2n\pi, n \in Z\] or

\[\frac{3x}{2} = 2m\pi + \frac{\pi}{2} - \frac{3x}{2}, m \in Z\]  (Taking negative sign will give absurd result.)

\[x = 2n\pi, n \in Z\] or 

\[x = \frac{2m\pi}{3} + \frac{\pi}{6}, m \in Z\]