Question

What are all the solutions to the equation in the interval (0,2pi)
Cos 4x - cos 2x = 0

Answer 1

you could use the trig identity
\[ \cos 2A = \cos^2 A - \sin^2 A = 2 \cos^2 A -1 \]
in other words, write \(\cos 4x\) as \( 2\cos^2 2x -1 \) to get the equation
\[ 2\cos^2 2x - \cos 2x -1 =0\]

Answer 2

\[\cos C-\cos D=-2\sin \frac{ C+D }{ 2 }\sin \frac{ C-D }{ 2 }\]
\[\cos 4x-\cos 2x=-2\sin \frac{ 4x+2x }{ 2 }\sin \frac{ 4x-2x }{ 2 }=-2\sin 3x \sin x\]
either\[ \sin3x=0=\sin n \pi,3x=n \pi,x=\frac{ n \pi }{ 3 },n=1,2,3,4,5\]
or \[\sin x=0=\sin npi,x=n \pi,n=1~ only.\]