Question

Solve the following equation:

\[\cos x \cos 2x \cos 3x = \frac{1}{4}\]

Answer 1

\[ \cos x \cos2x \cos3x = \frac{1}{4}\]
\[ \Rightarrow \left[ \frac{\cos\left( x + 2x \right) + \cos\left( 2x - x \right)}{2} \right]\cos3x = \frac{1}{4}\]
\[ \Rightarrow 2\left[ \cos3x + \cos x \right]\cos3x = 1\]
\[ \Rightarrow 2 \cos^2 3x + 2\cos x \cos3x - 1 = 0\]
\[ \Rightarrow 2 \cos^2 3x - 1 + 2\cos x \cos3x = 0\]
\[ \Rightarrow \cos6x + \cos4x + \cos2x = 0\]
\[ \Rightarrow \cos6x + \cos2x + \cos4x = 0\]
\[ \Rightarrow 2\cos4xcos2x + \cos4x = 0\]
\[ \Rightarrow \cos4x\left( 2\cos2x + 1 \right) = 0\]
\[ \Rightarrow \cos4x = 0 or 2\cos2x + 1 = 0\]
\[ \Rightarrow \cos4x = 0 or \cos2x = \frac{- 1}{2}\]
\[ \Rightarrow \cos4x = \cos\frac{\pi}{2} or \cos2x = \cos\frac{2\pi}{3}\]
\[ \Rightarrow 4x = \left( 2n + 1 \right)\frac{\pi}{2}, n \in Z or 2x = 2m\pi \pm \frac{2\pi}{3}, m \in Z\]
\[ \Rightarrow x = \left( 2n + 1 \right)\frac{\pi}{8}, n \in Z or x = m\pi \pm \frac{\pi}{3}, m \in Z\]