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Homework Help: Tough trig equation

  1. Mar 16, 2008 #1
    20sinx - 4cosx - 13 = 0 : solvable?

    Is this equation solvable? I've been banging my head on it for the past half hour and don't know how to start.

    20sinx - 4cosx - 13 = 0

    Also, this started out as a physics question, but seeing as this is just math stuff now I thought I'd put it here. I know all the physics was done correctly, so this is how it should end up.
    Last edited: Mar 16, 2008
  2. jcsd
  3. Mar 16, 2008 #2


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    Yes, you can solve it. Write sin(x)=sqrt(1-cos(x)^2). Now write the equation as 20sinx=4cosx+13 and square both sides. You should get a quadratic equation for cos(x). Solve it.
  4. Mar 16, 2008 #3
    Oh wow, that's brilliant!!! Thanks, Dick. I'm going to put that in my trig solving strategy log :)
  5. Mar 17, 2008 #4

    Gib Z

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    http://en.wikibooks.org/wiki/Trigonometry:Trigonometric_Identities_Reference [Broken]

    The first one in "Sum to product" should help in these questions =]
    Last edited by a moderator: May 3, 2017
  6. Mar 17, 2008 #5


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    It should be mentioned that the two methods, which are essentially different ways of making the same manipulation, will give you two answers: the first because it yields a quadratic equation, the second because equations of the form

    A sin x + B cos x = C

    can be transformed (using the identity GibZ refers to) into the form

    sin (x + phi) = K

    and can be solved if |K|<=1 .

    You will need to check the results against your original problem to see which solution (or perhaps both) applies to the situation.
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