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Another Trigonometric Equation

  1. Oct 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve for t

    sin3tcost+cos3tsint=-1/2

    2. Relevant equations

    Sum and Difference Identities

    3. The attempt at a solution

    I saw this problem the the sum identity sin([itex]\alpha[/itex]+[tex]\beta[/tex]) looked like the way to begin solving for t.

    This condenses to sin(3t+t)= -1/2 which is sin(4t)= -1/2

    I let 4t=x and now have sin x= -1/2

    The answers for this are 7[itex]\pi[/itex]/6+2[itex]\pi[/itex]n and 11[itex]\pi[/itex]+2[itex]\pi[/itex]n.

    This means that 4t equals both 7[itex]\pi[/itex]/6+2[itex]\pi[/itex]n and 11[itex]\pi[/itex]+2[itex]\pi[/itex]n. I need to divide the four out to get t by itself.

    The general solutions (I think) are t=7[itex]\pi[/itex]/24+[itex]\pi[/itex]n/2 and 11[itex]\pi[/itex]/6+[itex]\pi[/itex]n/2.

    Did I miss anything or does this look correct?

    Thanks!
     
  2. jcsd
  3. Oct 17, 2009 #2

    Mark44

    Staff: Mentor

    In the next couple of lines, the 2nd expression should be 11[itex]\pi[/itex]/6 +2[itex]\pi[/itex]n, which it appears you caught later.
    In the next line, you got the "divided by 6" back in the second expression, but forgot to divide by 4.
    Make the above t=7[itex]\pi[/itex]/24+[itex]\pi[/itex]n/2 and 11[itex]\pi[/itex]/24+[itex]\pi[/itex]n/2.
     
  4. Oct 17, 2009 #3
    You're right. I had that written on my paper, but it's a typo here on the board. Thank you for reading my post!
     
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