Question

sin 2x = sin x , 0 less then or equal to x less then or greater to 2pi. solve for x

Answer 1

oh ok ill take this one then..
use double angle identity, then factor

\[2\sin xcos x = \sin x\]
\[2\sin x \cos x - \sin x = 0\]
\[\sin x (2\cos x -1) = 0\]
set each factor = to 0 and solve for x

Answer 2

\[\sin 2x=\sin x,\sin 2x-\sin x=0,2\cos \frac{ 2x+x }{2 }\sin \frac{ 2x-x }{2 }=0\]
\[\cos \frac{ 3x }{2 }=0=\cos \frac{ \pi }{2 }=\cos \left( 2 n \pi+\frac{ \pi }{2} \right)\]
\[x=\frac{ 4n \pi }{3}+\frac{ \pi }{3 }\]
\[\sin x=0=\sin n \pi ,x=n \pi \]