Question

Find all solutions to the equation in the interval [0, 2π).

sin 2x - sin 4x = 0

Answer 1

sin(2x) = sin(4x)
sin(2x) = sin(2*2x)
sin(2x) = 2sin(2x)cos(2x)
2sin(2x)cos(2x) - sin(2x) = 0
sin(2x) * (2cos(2x) - 1) = 0
sin(2x) = 0 or cos(2x) = 1/2
x = 0 + n•π/2, π/6 + n•π, 5π/6 + n•π

x = 0, π/6, π/2, 5π/6, π, 7π/6, 3π/2, 11π/6

Answer 2

@FutureMathProfessor Why do you always give the answer away? That completely defeats the point of this website.

Answer 3

I showed him how to do it, I think you just need to cool it with the spam

Answer 4

Why did you bring in cos in step 3?

Answer 5

It's an identity

Answer 6

\[\bf \sin(2x)=2\sin(x)\cos(x) \implies \sin(2*2x)=sin(4x)=2\sin(2*x)\cos(2*x)\]Now re-write the equation:\[\bf \sin(2x)-2\sin(2x)\cos(2x)=0 \implies \sin(2x)(1-2\cos(2x))=0\]Can you find the x-values from here? @KirbyLegs

Answer 7

I can looking at a unit circle. I get the process all the way up to the last step.
Why is it 1-2cos(2x)

Answer 8

Oh because it's 2sin?

Answer 9

That's another identity

Answer 10

Ok thanks guys. I guess i just needed to pull out sin2x

Answer 11

Medal? :D