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How do i solve this trig equation

  • Thread starter Dell
  • Start date
  • #1
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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:

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
Try squaring both sides and see if you can use sin2x+cos2x=1 in hopes to get one trig term.
 
  • #3
590
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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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