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Homework Help: Trigonometric equation

  1. Dec 26, 2009 #1
    Hi,

    I would like som help on this one:

    Give all the real soulutions to the equation
    [tex]\frac{7}{4}-2sin(x) - cos^2(x) = 0[/tex]

    i tried the common identities, like replacing [tex]cos^2(x)[/tex] with [tex]1-sin^2(x)[/tex]

    any ideas?
     
  2. jcsd
  3. Dec 26, 2009 #2

    Mentallic

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    Well you picked the right identity. Why did you stop? :tongue:
    You'll have a quadratic in sinx which you need to solve, and remember that the range of sinx is between -1 and 1 so scrap any solutions outside this range as they're not real.
     
  4. Dec 26, 2009 #3

    tiny-tim

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    Hi James889! :smile:
    Good! :smile:

    and now put sinx = y, and solve. :wink:
     
  5. Dec 26, 2009 #4
    Thank you Flounder :),

    So something along the lines of:
    [tex]\frac{7}{4} -2sin(x)-cos^2(x) = 0[/tex]

    [tex]\frac{7}{4} -2sin(x)-1-sin^2(x) = 0 [/tex]

    [tex]\frac{3}{4}-2sin(x)-sin^2(x)=0 [/tex]

    substitute x = sin x

    [tex]x^2 -2x -\frac{3}{4} = 0[/tex]

    [tex]-\frac{1}{2} \pm \sqrt{\frac{4+4*\frac{3}{4}}{2}}[/tex]
     
  6. Dec 26, 2009 #5

    tiny-tim

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    not fishy enough

    nooo, not fishy enough :wink:

    i] you got a minus sign wrong

    ii] you still need to convert back

    iii] using the same letter to mean two different things ("substitute x = sin x") is a really mammalian thing to do. :rolleyes:
     
  7. Dec 26, 2009 #6
    Re: not fishy enough

    :frown:

    Hm, ye i think i messed up the quadratic formula i think it should be like this [tex]-x^2 -2x +\frac{3}{4} = 0[/tex]

    Giving me
    [tex]\frac{1}{2} \pm \sqrt{\frac{(-2)^2+4*\frac{3}{4}}{2}}[/tex]

    hm, convert back?
     
  8. Dec 26, 2009 #7

    tiny-tim

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    No, almost completely wrong. :redface:
    Yes, that'll be sinx, not x.
     
  9. Dec 27, 2009 #8
    Tim, tim, tim you're breaking my balls here...

    one more try...deep breath

    [tex]\frac{2}{2}\pm\sqrt{\bigg(\frac{-2}{2}\bigg)^2 -3/4}[/tex]

    So [tex]x = -0.5[/tex] but [tex]sin^{-1}(-0.5) = -30[/tex] that in the original equation gives 2
    but [tex]sin^{-1}(0.5) = 30[/tex] gives the correct answer of 0.
     
  10. Dec 27, 2009 #9

    tiny-tim

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    Mysteriously, that's the right answer, but I don't know how you got it from -x2 - 2x + 3/4 = 0 :confused:

    Nor how you get -0.5 from it :redface:
     
  11. Dec 27, 2009 #10

    Mentallic

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    Obviously the quadratic formula method isn't working too well...

    How about this?

    [tex]-x^2 -2x +\frac{3}{4} = 0[/tex]

    [tex]4x^2+8x-3=0[/tex] (by multiplying through by -4)

    Now, see if you can factorize this, because it can be factorized neatly.
     
  12. Dec 27, 2009 #11
    [tex](-2x+3)(2x-1)[/tex]

    But i don't get it, what was wrong with using the formula?
     
  13. Dec 27, 2009 #12

    tiny-tim

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    uhh? (-2x+3)(2x-1) = -4x2 + 8x - 3 :confused:

    (and your formula started with b/2a instead of -b/2a)
     
  14. Dec 27, 2009 #13

    Mentallic

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    As tiny-tim has said, you didn't factorize correctly. Try again and check your solution by expanding!

    You want:

    [tex]4x^2+8x-3[/tex]

    and you have:

    [tex](-2x+3)(2x-1)=-4x^2+8x-3[/tex]
     
  15. Dec 28, 2009 #14
    Hm,

    Is there some sort of trick when factorizing these kinds of expressions ?

    i have only gone by trial and error so far, this could take a while :/
     
  16. Dec 28, 2009 #15

    tiny-tim

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    I wouldn't bother to try to factor a quadratic equation unless the result is blindingly obvious …

    it's quicker and safer to use the (-b ± √etc) formula. :wink:
     
  17. Dec 28, 2009 #16
    I tried but it turned out fishy ;)

    yes, i had [tex] -x^2 -2x +\frac{3}{4} = 0[/tex]

    -b in this case would be [tex]\frac{-(-2)}{2}[/tex]
     
  18. Dec 28, 2009 #17

    tiny-tim

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    uhhh? :redface: what's wrong with being fishy? :frown:
    Yes, you've been very messy :rolleyes:, but you've got the right result at last: 1 ± 1/2. :smile:

    ok, so if sinx = 1 ± 1/2, x = … ?​
     
  19. Dec 28, 2009 #18
    [tex]x = 30^{\circ}[/tex]

    Is happy tim ? :)
     
  20. Dec 28, 2009 #19

    tiny-tim

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    i'm always happy! o:)
    mmm … so all the real solutions are … ? :smile:
     
  21. Dec 28, 2009 #20
    Oh, dang i forgot about that...
    [tex]\frac{\pi}{6} +n*2\pi[/tex]
     
    Last edited: Dec 28, 2009
  22. Dec 28, 2009 #21

    tiny-tim

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    oh come on

    you can't mix degrees and radians, can you?

    and anyway, 30º isn't the only solution ≤ 360º. :wink:
     
  23. Dec 28, 2009 #22
    Heh, i knew that:uhh: i was just, um testing you :rolleyes:

    of course, silly me, we have [tex]sin(\pi-\pi/6)[/tex] as a solution aswell.
     
  24. Dec 28, 2009 #23

    tiny-tim

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    ok, well i assume you're still testing me, so i think that would be :rolleyes:

    5π/6 :smile:
     
  25. Dec 28, 2009 #24

    Mentallic

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    Congrats tiny-tim, I think you passed the test! :wink:

    I'm just being a little picky here: when you say another solution is [itex]sin(5\pi/6)[/itex] remember that the original equation was [itex]-sin^2x-2sinx+3/4=0[/itex] and the solutions to this equation are x=... so don't say a solution is [itex]sin(5\pi/ 6)[/itex] but rather just [itex]5\pi /6[/itex] :smile:

    Else if you were to test your solution, you'd be solving this: [tex]-sin^2(sin(5\pi/ 6))-2sin(sin(5\pi/ 6))+3/4[/tex] which does not equal 0 thus it's not a solution.


    Ok so now you have the two solutions for [itex]0\leq x\leq 2\pi[/itex]. [itex]x=\pi/6[/itex] and [itex]x=5\pi/6[/itex]. Can you express all of the multiple values of these 2 solutions in just 1 equation? (I'm assuming that's how you're expected to show the values)

    e.g.

    [itex]x=\pi /6 +2\pi n[/itex]
    and
    [itex]x=5\pi /6 +2\pi n[/itex]
    n being all integers.

    Can you combine these 2 equations into 1?
     
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