Question

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

sin 2x - sin 4x = 0

Answer 1

I'm up to sin2x=2sin2xcos2x

Answer 2

good, now factor out the sin2x

--> sin2x (1 - 2cos2x) = 0

Answer 3

sin2x = 0 ----> 2x = 0,pi, x = 0,pi/2

1-2cos2x = 0
...

Answer 4

How do you know that 2x=0, π

Answer 5

and that x=0, π/2?

Answer 6

sin(angle) = 0 when angle is multiple of pi

angle = pi*n

2x = pi*n

x = (pi/2)*n

Answer 7

Okay, that makes sense

Answer 8

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

Answer 9

@dumbcow

Answer 10

@ganeshie8

Answer 11

Anyone please?

Answer 12

what i already did?

where did the sin come from, after factoring you have

sin2x = 0

1-2cos2x = 0

Answer 13

its simple algebra to isolate the "cos2x"

Answer 14

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

Answer 15

After factoring out I got sin2x(sin2xcos2x-1)

Answer 16

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

Answer 17

Oh okay thanks

Answer 18

I don't see it though