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Arctrig and Solving Trig. Equation

  1. Aug 24, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to find out the radian on the unit circle based on the questions below.
    1. tan3x=1
    2. 2sin(2t)+1=0

    2. Relevant equations



    3. The attempt at a solution
    1. tan3x=1
    tanx= (1/3) ?

    2. 2sin(2t)+1=0
    2sin(2t)= -1
    sin2t= (-1/2) ?
     
  2. jcsd
  3. Aug 24, 2009 #2

    rock.freak667

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    1) No (do what you did in question 2)

    2) Yes.


    continue on now
     
  4. Aug 24, 2009 #3
    I didn't really know what to do after that.
    Could you give me a hint.
     
  5. Aug 24, 2009 #4

    rock.freak667

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    if tanA=x, then A=tan-1(x). This is similar for sin and cos.
     
  6. Aug 24, 2009 #5
    Thank you so much. It did help me a lot.
     
  7. Aug 24, 2009 #6

    VietDao29

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    If you want to find the general solution, then you may want to memorize these formulae:

    [tex]\sin(x) = y \Leftrightarrow \left[ \begin{array}{ccc} x & = & \arcsin(y) + 2k \pi \\ x & = & \pi - \arcsin(y) + 2k' \pi \end{array} \right. , k , k' \in \mathbb{Z}[/tex]

    Since sin has the period of [tex]2 \pi[/tex], and [tex]\sin(\pi - x) = \sin(x)[/tex].

    [tex]\cos(x) = y \Leftrightarrow \pm \arccos(y) + 2k \pi , k \in \mathbb{Z}[/tex]

    Since cos has the period of [tex]2 \pi[/tex], and [tex]\cos(- x) = \cos(x)[/tex].

    [tex]\tan(x) = y \Leftrightarrow \arctan(y) + k \pi , k \in \mathbb{Z}[/tex]

    Since tan has the period of [tex]\pi[/tex].

    [tex]\cot(x) = y \Leftrightarrow \mbox{arccot}(y) + k \pi , k \in \mathbb{Z}[/tex]

    Since cot has the period of [tex]\pi[/tex].
     
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