Question

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

sin2x - sin4x = 0

Answer 1

sin4x can be written : 2sin(2x)cos(2x)
does it help ?

Answer 2

kind of. i'm still a bit confused though. so what i have so far is
2sinxcosx - 2sin(2x)cos(2x) = 0

Answer 3

would i find a gcf next and factor out the 2sinxcosx?

Answer 4

i think it is better to write it :
sin(2x) - 2sin(2x)cos(2x)=0
and now you can factor sin(2x) out

Answer 5

sin(2x) * (1-2cos(2x)) = 0
you have to find
the values:
sin(2x) = 0
1-2cos(2x) = 0

Answer 6

I don't know if you are familiar with this identity or not but if you are you can use:
\[\sin(x)-\sin(y)=2 \cos(\frac{x+y}{2})\sin(\frac{x-y}{2})\]

Answer 7

and if you don't @Coolsector has a cool style too :)

Answer 8

@freckles i haven't learned about sum to product formulas yet so i'm not familiar with that. i just learned about multiple angle formulas and sum and difference formulas so that's what i'm trying to use but thanks though!

Answer 9

and @Coolsector thank you that helped!

Answer 10

yw