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Solving trigonometric equations with a specific range

  1. Jul 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Given that -180 <= x <= 180, solve the following equation in radians
    cosx = -1/sqrt(2)

    3. The attempt at a solution

    I'm not exactly sure how to solve this question with respect to the range -180 <= x <= 180. I know that -1/sqrt(2) means that x is 45 degrees in either the 2nd or 3rd quadrant but I do not know how to proceed from there.
     
  2. jcsd
  3. Jul 27, 2013 #2

    haruspex

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    Quite so. Can you write the complete sets of possible solutions in terms of an unknown integer n?
     
  4. Jul 27, 2013 #3

    Simon Bridge

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    I thought the 3rd quadrant was between 180deg and 270deg? Isn't that outside the range? ;)
    Of course, taking x as cyclic, then +180 to +270 is also -90 to -180.
    What is the value of x when the angle is "45 degrees in the second quadrant"?

    I'd have suggested sketching the cosine function for the range... ##-\pi \leq x \leq \pi##
    (since the answer should be in radians....) but I'm interested in where haruspex is going.
     
  5. Jul 27, 2013 #4

    QuantumCurt

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    If you start at -180, then trace the unit circle counter clockwise, what's the first point you get to that satisfies the equation, while still staying in the range of being less than 180 degrees? This is kind of a confusing range, because -180 and 180 exist at the same point in the unit circle, but there is a value between the two that satisfies.
     
  6. Jul 28, 2013 #5

    ehild

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    Draw the unit circle and see what it means "45 degrees in the second or third quadrant?" 45 with respect to what?
    Remember, positive angles are measured from the positive horizontal axis anti-clockwise, the negative angles are measured clockwise.

    ehild
     

    Attached Files:

  7. Jul 28, 2013 #6

    haruspex

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    Well, it's also between 540 degrees and 630 degrees, and between 900 degrees and 990 degrees, .... Unless you want call those the 7th and 11th quadrants, etc.
     
  8. Jul 28, 2013 #7

    Simon Bridge

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    ... etc etc etc :)
     
  9. Jul 28, 2013 #8
    Well if I start from -180 and proceed clockwise how is it possible for me to ever reach 180 because the most clockwise I go I will just be going into more negative numbers -180, -270, -360. Also the way I got 45 degrees in a either the 2nd or 3rd quadrant is by using special triangles and the cast rule.
     
  10. Jul 28, 2013 #9

    QuantumCurt

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    You need to start at -180, and proceed COUNTERclockwise. The range states -180<x<180

    Start at -180, and proceed in the direction that is going to cover the values that are greater than -180, which is counter clockwise. Remember that you're dealing in negative values. For example, 270 degrees on the unit circle is the same as -90 degrees. Which is greater than -180. Then you just keep going until you find a value(s) that satisfies the equation, while still being less than positive 180 degrees.
     
    Last edited: Jul 28, 2013
  11. Jul 28, 2013 #10

    QuantumCurt

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    Last edited: Jul 28, 2013
  12. Jul 28, 2013 #11
    Oh the interactive circle makes it so much more clearer. Essentially there are two solutions to this problem but can only be achieved through a positive angle and a negative angle because for example the two angle I would get would be 135 and 225 but since 225 is greater than 180 I have to rewrite it as -135 or 3pi/4 and -3pi/4. Many thanks to all!
     
  13. Jul 28, 2013 #12

    QuantumCurt

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    You got it. You're welcome. :)
     
  14. Jul 28, 2013 #13

    haruspex

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    I'll spell out where I was going with this since I feel it may help in more complex variations.
    If θ = ψ is one solution then θ = ψ + 2πn is also a solution for any integer n. So here you have a third quadrant solution, 5π/4, and want a corresponding solution in (-π, +π]. Solve -π < 5π/4 + 2πn <= +π: -4 < 5 + 8n <= 4; -9 < 8n <= -1; n = -1.
     
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