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Homework Help: Trig Equation

  1. Oct 3, 2005 #1
    3 cos x + 4 = 0
    cos x = -4/3

    How do I solve this? I can't take the arcsine...

    ---

    sin (3x - 40) = 0

    Is the general solution 73.3 + 120k | k E I?
     
  2. jcsd
  3. Oct 3, 2005 #2

    robphy

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    Did you try to use complex numbers or hyperbolic trig functions?
     
  4. Oct 3, 2005 #3

    hotvette

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    Could try graphing it and see where it crosses zero.
     
  5. Oct 3, 2005 #4
    It's not going to cross the x-axis. You'll have to encorporte complex numbers like robphy said.
     
  6. Oct 3, 2005 #5
    We don't cover complex numbers until the end of the year. Does this mean I say it has no solution?
     
  7. Oct 3, 2005 #6
    Why can you not take the cosine inverse of x?
     
  8. Oct 3, 2005 #7
    4/3 > 1 so you get a domain error... or at least that's how I see it.
     
  9. Oct 3, 2005 #8
    Unless I'm missing something, your given value for cos x is incorrect. Any value of the cosine function is in a way "stuck" between -1 and 1.

    Use a calculator and take the cosine of any degree/radian measure you wish, it will always be between -1 and 1.
     
    Last edited: Oct 4, 2005
  10. Oct 3, 2005 #9
    I got -4/3 from rearranging the equation. The fact that 4/3 > 1 is my problem.
     
  11. Oct 3, 2005 #10
    Is this an equation that needs solving, or an identity that needs verifying?
     
  12. Oct 3, 2005 #11
    "Solve for all possible values of x."
     
  13. Oct 3, 2005 #12
    no solution
     
  14. Oct 3, 2005 #13
    Thank you.
     
  15. Oct 4, 2005 #14
    I have another problem :frown:

    sin^2 x = 3/4

    for one of the possible answers I get [itex]\frac{\pi}{3} + 2\pi k[/itex] but my book says it should be [itex]\pi[/itex] instead of [itex]2\pi[/itex] for the period. How come?
     
  16. Oct 4, 2005 #15
    The sin value is squared, so the negative values square out to give a solution too.
     
  17. Oct 4, 2005 #16
    Yes I know. Let me clarify: [itex]\frac{\pi}{3} + 2\pi k[/itex] is one possible solution but my book says it should be [itex]\frac{\pi}{3} + \pi k[/itex] and I don't know why.
     
  18. Oct 4, 2005 #17

    robphy

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    [tex]e^{i\pi}=-1[/tex]
     
  19. Oct 4, 2005 #18
    I haven't done complex numbers so I don't really know the significance of that expression.
     
  20. Oct 4, 2005 #19

    TD

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    [tex]\sin ^2 x = \frac{3}
    {4} \Leftrightarrow \sin x = \pm \sqrt {\frac{3}
    {4}} = \pm \frac{{\sqrt 3 }}
    {2}[/tex]

    Now determine the solution in both cases (the + and the - case).
     
  21. Oct 4, 2005 #20
    I would say

    [tex]\frac{\pi}{3} + 2\pi k | k \epsilon I, \frac{4\pi}{3} + 2\pi k | k \epsilon I[/tex]

    but my book says

    [tex]\frac{\pi}{3} + \pi k | k \epsilon I, \frac{4\pi}{3} + \pi k | k \epsilon I[/tex]

    and I don't know why
     
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