Problems

Solution: \begin{align*} && \cos(A \pm B) &= \cos A \cos B \mp \sin A \sin B \\ A = a, B = 3a&& \cos 4a + \cos 2a &= 2\cos 3a \cos a \\ \Rightarrow && \cos a \cos 3a &= \tfrac12(\cos 4a + \cos 2a) \\ \\ && \sin(A \pm B) &= \sin A \cos B \pm \cos A \sin B \\ && \sin 4a + \sin(- 2a) &= 2 \sin a \cos 3a \\ \Rightarrow && \sin a \cos 3a &= \tfrac12 (\sin 4a - \sin 2a) \end{align*}

1. \(\,\) \begin{align*} && 1 &= 4 \cos x \cos 2x \cos 3x \\ &&&= 2(\cos 4x +\cos 2x)\cos 2x \\ c = \cos 2x:&&&= 2(2c^2-1+c)c \\ \Rightarrow && 0 &= 4c^3+2c^2-2c-1 \\ &&&= (2c+1)(2c^2-1) \\ \Rightarrow && \cos 2x &= -\frac12 \\ \Rightarrow && x &= \frac{\pi}{3}, \frac{2\pi}{3} \\ && \cos 2x &= \pm \frac1{\sqrt2} \\ \Rightarrow && x&= \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \end{align*}
2. \(\,\) \begin{align*} && \tan x &= \tan 2x \tan 3x \tan 4x \\ \Rightarrow &&1 &= \frac{\cos x\sin 2x \sin 3x \sin 4x}{\sin x \cos 2x \cos 3x \cos 4x} \\ &&&= \frac{\sin 2x \sin 4x (\sin4 x + \sin 2x)}{\cos 2x \cos 4x (\sin 4x - \sin 2x)} \\ &&&= \frac{(\cos 2x - \cos 6x) (\sin4 x + \sin 2x)}{(\cos 6x + \cos 2x) (\sin 4x - \sin 2x)} \\ \Rightarrow && 0 &= 2\cos 6x \sin 4x - 2\cos 2x \sin 2 x\\ &&&= \sin 4 x (2 \cos 6x - 1) \\ \Rightarrow && \sin 4x &= 0 \\ \text{ or }&& \cos 6x &= \frac12 \end{align*} \(\sin 4x = 0 \Rightarrow x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi\) \(\cos 6x = \frac12 \Rightarrow x = \frac{\pi}{18}, \frac{5\pi}{18},\frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}\). We should verify these work, since not all of them will, especially where \(\sin 4x = 0\), so our final answer is \(x = 0, \pi, \frac{\pi}{18}, \frac{5\pi}{18},\frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}\)