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

sin 2x - sin 4x = 0

Question

Find all solutions to the equation in the interval [0, 2π).

sin 2x - sin 4x = 0

anonymous · Answer

I'm up to sin2x=2sin2xcos2x

dumbcow · Answer

good, now factor out the sin2x

--> sin2x (1 - 2cos2x) = 0

dumbcow · Answer

sin2x = 0   ----> 2x = 0,pi,  x = 0,pi/2

1-2cos2x = 0
...

anonymous · Answer

How do you know that 2x=0, π

anonymous · Answer

and that x=0, π/2?

dumbcow · Answer

sin(angle) = 0  when angle is multiple of pi

angle = pi*n

2x = pi*n

x = (pi/2)*n

anonymous · Answer

Okay, that makes sense

anonymous · Answer

How do I solve for the sin2xcos2x-1=0?

anonymous · Answer

@dumbcow

anonymous · Answer

@ganeshie8

anonymous · Answer

Anyone please?

dumbcow · Answer

what i already did?

where did the sin come from, after factoring you have

sin2x = 0

1-2cos2x = 0

dumbcow · Answer

its simple algebra to isolate the "cos2x"

anonymous · Answer

Wait I thought when you factored out the sin2x there's 1 sin2x left in the 2sin2x

anonymous · Answer

After factoring out I got sin2x(sin2xcos2x-1)

dumbcow · Answer

no that is wrong, if you multiply that out you will have a sin^2

anonymous · Answer

Oh okay thanks

anonymous · Answer

I don't see it though