Question

Solve the following equation:

\[\cos x + \cos 3x - \cos 2x = 0\]

Answer 1

\[\cos x + \cos 3x - \cos 2x = 0\]
\[\Rightarrow 2 \cos \left( \frac{4x}{2} \right) \cos \left( \frac{2x}{2} \right) - \cos2x = 0\]
\[ \Rightarrow 2 \cos2x \cos x - \cos2x = 0\]
\[ \Rightarrow \cos2x ( 2 \cos x - 1) = 0\]

\[\Rightarrow \cos2x = 0\] or

\[2 \cos x - 1 = 0\]

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

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

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

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

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

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