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Trig Equation #2

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve for x on the interval 0<= theta <= 2pi. Give exact solutions if possible. Otherwise round to the nearest hundredth of a radian.

    2. Relevant equations
    tan^2x-4tanx = 0

    3. The attempt at a solution

    Tried factoring it to (tanx-4)(tanx) = 0

    tanx=4 or 0 I guess? Not sure where to go from there

    or instead of factoring I could immediately make it sin/cos, which gives

    (sin^2x - 4sincos) / (cos^2x) = 0
    Last edited: Nov 24, 2008
  2. jcsd
  3. Nov 24, 2008 #2


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    I would plot a graph of tan(x) for [itex]x \in [0,\pi][/itex]....at what points does tanx=4?...At what points does tanx=0?...Those will be your solutions.
  4. Nov 24, 2008 #3
    "Tried factoring it to (tanx-4)(tanx) = 0"

    @Draggu now it is good. So as you said there are two solutions: tanx=4 and tanx=0.

    The solutions for x are:



  5. Nov 24, 2008 #4
    We have not learned that yet in class... so doing that wouldn't work. :p
  6. Nov 24, 2008 #5
    There is no exact radian measure/degree where tanx=4
    Tanx=4 approx at 76degrees. God I'm lost.
  7. Nov 24, 2008 #6
    Well I got: (19pi/45), (64pi/45), pi, 2pi as the answers
  8. Nov 25, 2008 #7
    You know how to draw graphic of tan function, and do not know how to find arctan(x)?

    That's strange.

    Anyway, you got the solutions correctly, since the task is to find the solutions in the interval [0,2п]. In the future, you will learn about periods.So if you want to be more correctly :smile: you should write 19п/45 + kп and п + kп, where k Є Z
  9. Nov 25, 2008 #8
    Well, for tan(x) = 0, you don't need to know tan(x). [tex]tan(x) = \frac{sin(x)}{cos(x)}[/tex] so tan(x) = 0 when sin(x) = 0.
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