Question

Please show me how to do this!
find all SOLUTIONS of the following equation in the interval [0,2π)

cosx=cos3x

Answer 1

\[\cos x=\cos 3x\]
cosx-cos3x=0
\[\cos C-\cos D=2\sin \frac{ C+D }{ 2 }\sin \frac{ D-C }{ 2 }\]
use and solve

Answer 2

\[2\sin \frac{ x+3x }{ 2 }\sin \frac{ 3x-x }{ 2 }=0\]
\[\sin 2x \sin x=0\]
\[either \sin 2x=0=\sin n \pi \]
\[2x=n \pi,n=0,1,2,3,4\]
\[x=\frac{ n \pi }{ 2 }\]
\[x=0,\frac{ \pi }{ 2 },\pi,\frac{ 3\pi }{ 2 },2\pi \]
\[or \sin x=0=\sin n \pi ,x=n \pi,n=0,1,2\]
\[x=0,\pi,2\pi\]
\[hen ce x=0,\frac{ \pi }{ 2 },\pi,\frac{ 3\pi }{ 2 },2\pi \]