\input amstex

$$\aligned
\cos 2x-\cos 4x+\cos 6x&=0\cr
t&=2x\cr
\cos t-\cos 2t+\cos 3t&=0\cr
\cos^3t-3\cos t\sin^2t-\cos^2t+\sin^2t+\cos t&=0\cr
4\cos^3t-2\cos^2t-2\cos t+1&=0\cr
4\cos^32x-2\cos^22x-2\cos 2x+1&=0\cr
2\cos2x\,(2\cos^22x-1)-(2\cos^2x-1)&=0\cr
(2\cos2x-1)(2\cos^22x-1)&=0\cr
\Bigl(\cos2x=\frac{1}{2}\Bigr)&\lor
\Bigl(|\cos2x|=\frac{1}{\sqrt{2}}\Bigr)\cr
\Bigl(\pm2x=\arccos\frac{1}{2}\Bigr)&\lor
\biggl(\pm2x=\arccos\Bigl(\pm\frac{1}{\sqrt{2}}\Bigr)\biggr)\cr
\Bigl(x=\pm\frac{\pi}{6}\Bigr)&\lor\biggl(x\in\Bigl\{\pm\frac{\pi}{8},
\pm\frac{3}{8}\pi\Bigr\}\biggr)\cr
\cos\Bigl(\frac{\pi}{3}\Bigr)=\cos\Bigl(-\frac{\pi}{3}\Bigr)%
&=\cos\Bigl(-\frac{\pi}{3}+2\pi\Bigr)=\cos\Bigl(\frac{5}{3}\pi\Bigr)\cr
\cos\Bigl(2\cdot\frac{\pi}{6}\Bigr)=\cos\biggl(2\cdot
\Bigl(-\frac{\pi}{6}\Bigr)\biggr)&=\cos\Bigl(2\cdot\frac{5}{6}\pi\Bigr)%
\;\Longrightarrow\;\{x\}\ni\pm\frac{5}{6}\pi\cr
\cos\Bigl(\frac{\pi}{4}\Bigr)=\cos\Bigl(-\frac{\pi}{4}\Bigr)%
&=\cos\Bigl(-\frac{\pi}{4}+2\pi\Bigr)=\cos\Bigl(\frac{7}{4}\pi\Bigr)\cr
\cos\Bigl(2\cdot\frac{\pi}{8}\Bigr)=\cos\biggl(2\cdot
\Bigl(-\frac{\pi}{8}\Bigr)\biggr)&=\cos\Bigl(2\cdot\frac{7}{8}\pi\Bigr)%
\;\Longrightarrow\;\{x\}\ni\pm\frac{7}{8}\pi\cr
\cos\Bigl(\frac{3}{4}\pi\Bigr)=\cos\Bigl(-\frac{3}{4}\pi\Bigr)%
&=\cos\Bigl(-\frac{3}{4}\pi+2\pi\Bigr)=\cos\Bigl(\frac{5}{4}\pi\Bigr)\cr
\cos\Bigl(2\cdot\frac{3}{8}\pi\Bigr)=\cos\biggl(2\cdot
\Bigl(-\frac{3}{8}\pi\Bigr)\biggr)&=\cos\Bigl(2\cdot\frac{5}{8}\pi\Bigr)%
\;\Longrightarrow\;\{x\}\ni\pm\frac{5}{8}\pi\cr
x&\in\Bigl\{\pm\frac{\pi}{8}, \pm\frac{\pi}{6}, \pm\frac{3}{8}\pi,
\pm\frac{5}{8}\pi, \pm\frac{5}{6}\pi, \pm\frac{7}{8}\pi\Bigr\}
\endaligned$$
\bye