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How do i solve this trig equation

  1. Apr 28, 2009 #1
    as part of a physics problem i get to one stage before the ansewr and then i get stuck here:

    0.15348=0.1415cosβ -0.291sinβcosβ

    how do i solve this equation with both sinβ and cosβ, i realise that i need to play with the identities but have had no luck,
    please help
     
    Last edited: Apr 28, 2009
  2. jcsd
  3. Apr 28, 2009 #2

    rock.freak667

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    Homework Helper

    Try squaring both sides and see if you can use sin2x+cos2x=1 in hopes to get one trig term.
     
  4. Apr 28, 2009 #3
    copied it all wrong, sorry, obviously cant work,,,

    meant to be

    0.15348=0.1415cosβ -0.291sinβcosβ

    if i square it i get

    0.023556=0.02cos2β +0.085sin2βcos2β-0.08sinβcosβ

    if i say cos2β=t

    0.023556=0.02t + 0.085(1-t)*(t)-0.08*[tex]\sqrt{1-t}[/tex][tex]\sqrt{t}[/tex]

    0.023556=0.02t + 0.085t - 0.085t2 -0.08[tex]\sqrt{t-t2}[/tex]

    0.023556=0.105t - 0.85t2 - 0.08[tex]\sqrt{t-t2}[/tex]

    now how would i find t??

    you sure theres no better way?
     
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