Question

I'm really confused on how to solve this. Anyone able to help?

sin4x = -2sin2x

Answer 1

\[\sin 4x=-2\sin 2x\]
\[2\sin 2x \cos 2x+2\sin 2x=0\]
\[2\sin 2x \left( \cos 2x+1 \right)=0\]
\[Either~\sin 2x=0=\sin n \pi,2x=n \pi,x=n \frac{ \pi }{ 2 }\]
or\[\cos 2x=-1=\cos \left( 2n+1 \right)\pi,x=\left( 2n+1 \right)\frac{ \pi }{ 2 }\]
here n is an integer.

Answer 2

I have to do it between 0 & 2pi, so it would be \[\frac{ \pi }{ 2 }\]