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Solving Trigonometry Equations

  1. Aug 4, 2012 #1
    1. The problem statement, all variables and given/known data

    My trig is really rusty and I've been trying to figure out why the answer is what it is:
    Find all solutions of the following equation:
    sin2 x + cos x - 1.

    2. Relevant equations

    Just the identity:
    sin2 θ + cos2 θ = 1

    3. The attempt at a solution

    The sin2 x is in the way, so I substitute 1 - cos2 x in it's place to get:
    (1 - cos2 x + cos x - 1 = 0

    The ones are removed and cos x is common, so:
    cos x(-cos x + 1) = 0

    This means that:
    cos x = 0 or cos x = -1

    From the unit circle, we get:

    cos x = ∏/2 or cos = ∏

    Here is where I get confused:

    I know we need to make this true for all intervals so:

    cos x = 0 whenever x = ± ∏/2 + 2k∏ or,
    cos x = -1 whenever x = ± ∏ + 2k∏ for any integer k.

    That's my final answer. According to my textbook, It's wrong, but I have no idea why. The textbook gives the final answer as:

    cos x = 0 whenever x = ∏/2 + 2k∏ or,
    cos x = -1 whenever x = ∏ + 2k∏ for any integer k.

    Is it wrong that x = ± ∏/2 or x = ±∏?
     
  2. jcsd
  3. Aug 4, 2012 #2

    ehild

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    Check the piece in red.

    By the way, x = ± ∏ + 2k∏ means the same x values as x = ∏ + 2k∏.

    If you take n=k-1 then - ∏ + 2k∏=∏+2(k-1)∏=∏+2n∏.


    ehild
     
  4. Aug 4, 2012 #3
    Thanks for the correction.
    I see now. It's not wrong, just redundant.

    Thanks a lot for your help.
     
  5. Aug 4, 2012 #4

    ehild

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    Yes, but the solution is not correct. -cosx+1=0 does not mean cosx=-1.

    ehild
     
  6. Aug 4, 2012 #5
    Yes. - cos x + 1 = 0 is supposed to be cos x = 1, right?
     
  7. Aug 4, 2012 #6

    ehild

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    Yes. So x=?

    ehild
     
  8. Aug 7, 2012 #7
    Sorry for the late reply. Things have been hectic.

    x = 0, I think.
     
  9. Aug 7, 2012 #8

    Mentallic

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    Come on, you're learning to provide all the solutions here! What else does x equal?
     
  10. Aug 8, 2012 #9
    x = 0 whenever x = ∏/2 + 2∏k or
    x = 1 whenever x = 2∏ + 2∏k for any integer k.

    Was there a mistake in the textbook when it said cos x = -1 whenever x = ∏ + 2k∏ for any integer k.
     
    Last edited: Aug 8, 2012
  11. Aug 8, 2012 #10

    Mentallic

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    I believe you were meant to say
    [itex]\cos(x)=0[/itex] whenever x=...
    as opposed to x=0.

    Now, you're nearly correct. Take a look at the graph of [itex]y=\cos(x)[/itex] and you should notice that while it touches its extreme values of -1 and 1 only once every [itex]2\pi[/itex] units, it cuts everything else in between twice every [itex]2\pi[/itex] units. It cuts the y-axis (or cos(x)=0) every [itex]\pi[/itex] units, so your answer should be

    [itex]\cos(x)=0[/itex] whenever [itex]x=\frac{\pi}{2}+\pi k[/itex] which you can also represent (and should become familiar with) the factorized form [itex]x=\pi(\frac{1}{2}+k)[/itex]

    Also, for [itex]\cos(x)=1[/itex], your answer of [itex]x=2\pi+2\pi k[/itex] is a convoluted way of saying [itex]x=2\pi k[/itex] which is the same thing. Do you see why?
     
  12. Aug 8, 2012 #11
    Yes, I did mean cos (x) = 0 whenever x = ...

    I see it now. Trig has never been my strong point and I'll make sure to practise it some more, especially the basics.

    Thanks a whole lot for being so patient Mentallic and ehild.
     
  13. Aug 8, 2012 #12
    Yes, I did mean cos (x) = 0 whenever x = ...

    I see it now. Trig has never been my strong point and I'll make sure to practise it some more, especially the basics.

    Thanks a whole lot for being so patient Mentallic and ehild.
     
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