1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Challenging Trig Equation

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    2sin(4x)-sin(2x)-(√3)cos(2x)=0

    x is [0,2π]
    2. Relevant equations



    3. The attempt at a solution
    Using trig identities:
    8sinxcos3x-6sinxcosx-√3(2cos2x-1)

    Please help me with this problem. I have no idea where to go from here. Thanks.
     
    Last edited: Feb 15, 2012
  2. jcsd
  3. Feb 15, 2012 #2
    I've further simplified it to:

    4sin2x−tan2x−√3=0

    Does this make it easier? Please help.
     
    Last edited: Feb 15, 2012
  4. Feb 16, 2012 #3

    Mentallic

    User Avatar
    Homework Helper

    I was going to give this a try but had to head off to work, but now that I'm looking at it I'm honestly stuck as well...

    This doesn't seem to be making things simpler.

    I got here also and couldn't do much with it.

    I also converted [tex]\sin(2x)+\sqrt{3}\cos(2x)[/tex] into the [tex]R\sin(2x+\theta)[/tex] form and got [tex]2\sin(2x+\frac{\pi}{3})[/tex]

    so what needs to be solved now is

    [tex]\sin(4x)=\sin(2x+\frac{\pi}{3})[/tex]

    But I'm kind of stuck here as well.

    We'll get back to you!
     
  5. Feb 16, 2012 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    sin(4x)=sin(2x+π/3)

    When sin(α)=sin(β) either

    α=β+2kπ

    or

    α=(π-β)+2kπ.

    ehild
     
    Last edited: Feb 16, 2012
  6. Feb 16, 2012 #5

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    First, make everything in the angle 2x. This gives (as you already know)

    [tex]4\sin(2x)\cos(2x)-\sin(2x)-(√3)\cos(2x)=0[/tex]

    Now the trick is to use Weierstrass substitution (which is a handy technique in integration but it works here too).

    Let [itex]t=\tan(x)[/itex], then

    [tex]\sin(2x)=\frac{2t}{1+t^2},~\cos(2x)=\frac{1-t^2}{1+t^2}[/tex]

    this will give a polynomial in t which should be easier to solve.

    Not saying that this is the easiest way, but it gives a solution nonetheless.
     
  7. Feb 17, 2012 #6
    Thanks guys :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Challenging Trig Equation
  1. Trig equations (Replies: 6)

  2. Trig equation (Replies: 1)

  3. Trig equation (Replies: 4)

  4. Trig. Equation (Replies: 8)

  5. Trig equation (Replies: 10)

Loading...