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Solving Trigonometric Equations

  1. Apr 5, 2007 #1
    1. The problem statement, all variables and given/known data
    use the inverse functions where necessary to find all solution of the equation in the interval [0,2pi). Use a graphing utility to verify.


    2. Relevant equations

    3. The attempt at a solution

    The answer is pi/2 and 3pi/2 from my teacher..

    stuck :confused:
  2. jcsd
  3. Apr 5, 2007 #2
    You can solve it like any other quadratic. What would you do with 2y2 + 3y = 0 ?
  4. Apr 6, 2007 #3


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    also remember you cannot divide both side by cos x (or y in turdferguson redefinition), because cos x can be zero (and in fact it is a solution)
  5. Apr 6, 2007 #4


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    As turdferguson points out, it's not completely valid to divide both sides by cos(x). What if cos(x)=0?
  6. Apr 6, 2007 #5
    I can factor out cosx? (sorry, this is midterm, and this stuff is what we did in January:p)

    cosx=0 and cosx=-3/2

    I can figure cosx=0 to be x=pi/2 and 3pi/2 but can anyone remind me why cosx=-3/2 doesn't work (invalid in calc)?
    I think because adj/hyp, and adj has to be smaller than hyp, so cos or sin > 1 = no solution?
  7. Apr 6, 2007 #6


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    Yes, that's the reason.
  8. Apr 6, 2007 #7


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    No real number solution. It is possible to have a complex value that has cosine equal to 1.5.
  9. Apr 6, 2007 #8
    Could you also look at it this way? f(x)=2cox(x)+3 -- It's going to be translated? up 3 units and is only has an amplitude of 2 so it will never cross the x-axis, so there are not any x-intercepts.
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