anonymous
  • anonymous
Find all solutions to the equation in the interval [0, 2π). sin 2x - sin 4x = 0 I NEED HELP I DON'T UNDERSTAND THIS CRAP.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
srry dont really know
anonymous
  • anonymous
im only in geomerty
anonymous
  • anonymous
how helpful of you!

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anonymous
  • anonymous
is it multipulchoise
Lukecrayonz
  • Lukecrayonz
sin(2x) + sin(4x) = 0 sin(2x) + 2sin(2x)cos(2x) = 0 sin(2x)[1 + 2cos(2x)] = 0 sin(2x) = 0 or 2cos(2x) = -1 -> cos(2x) = -1/2 when sin(2x) = 0 2x = 0, pi, 2pi, 3pi x = 0, pi/2, pi, 3pi/2 when cos(2x) = -1/2 2x = 2pi/3, 4pi/3, 8pi/3, 10pi/3 x = pi/3, 2pi/3, 4pi/3, 5pi/3 so x = 0, pi/2, pi, 3pi/2 , pi/3, 2pi/3, 4pi/3, 5pi/3 @jim_thompson5910 is thsi correct?
anonymous
  • anonymous
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, 2π
anonymous
  • anonymous
As the king I demand you take your sluttish Profile Picture off and put something more holy. What would your parents say.
anonymous
  • anonymous
a. π/6 , π/2 , 5π/6 , 7π/6 , 3π/2, 11π/6 b. 0, π/6 , π/2 , 5π/6 , π, 7π/6 , 3π/2, 11π/6 c. 0, 2π/3, 4π/3 d. 0, π/3, 2π/3, π, 4π/3, 5π/3
anonymous
  • anonymous
as reaper i sujuest u shh or die
anonymous
  • anonymous
My lord, You people can't take a joke.
anonymous
  • anonymous
i was jst kiddn
anonymous
  • anonymous
@AFleming42 dose that help u at all??
anonymous
  • anonymous
Oh, I wasn't that is a very sluttish picture. Parents you watch what there kids wear these days.
anonymous
  • anonymous
that was a graduation party
anonymous
  • anonymous
How did you get that answer @Reaper534 ?
anonymous
  • anonymous
by the power of google hahaha
anonymous
  • anonymous
Alright well I'll take google's word on it and hope for the best!
anonymous
  • anonymous
thanks mucho(:
anonymous
  • anonymous
Your welcome :)
anonymous
  • anonymous
Glad i could help.
jim_thompson5910
  • jim_thompson5910
Lukecrayonz you are correct, so that would make the answer choice B
anonymous
  • anonymous
In that case thank you again @jim_thompson5910 !
jim_thompson5910
  • jim_thompson5910
didn't do much, Lukecrayonz did all the work really
anonymous
  • anonymous
good job @Lukecrayonz (: <3
Lukecrayonz
  • Lukecrayonz
thanks cutie can i get your number
anonymous
  • anonymous
message me(;
anonymous
  • anonymous
ernies back

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