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Homework Help: Trig Equation

  1. May 2, 2006 #1
    Hey,

    So... I have this problem:

    -sin(x)-2cos(2x)

    And I need to solve for all the x values between 0 and 2pi.


    Naturally, one would equate that to zero and get:

    -sin(x)-2cos(2x)=0

    And in order to find what that's equivalent to, you'd have to find a way to factor it... And in order to do that, you'd have to use identities... I know the following for cos(2x):

    [tex]\cos(2x)=\cos(x)^2-\sin(x)^2[/tex]
    [tex]\cos(2x)=2\cos(x)^2-1[/tex]
    [tex]\cos(2x)=1-2\sin(x)^2[/tex]


    However, I don't see any of those actually helping. By using the last one, I get:

    [tex]4\sin(x)^2-\sin(x)-2[/tex]

    But there's no reasonable way to factor that. I went through the depths of complete the square hell to find this:

    [tex]\frac{1}{2}(2\sin(x)-1)(4\sin(x)-1)-\frac{3}{2}=0[/tex]
    [tex](2\sin(x)-1)(4\sin(x)-1)=3[/tex]

    And if I set each parenthesis equal to 3, which I'm pretty sure you can't do, I get:

    [tex]\arcsin(2)[/tex] and [tex]\arcsin(1/2)[/tex]

    Both of which are wrong...


    Can anyone shed any light on this?
    Haha, I'm having quite a bit of difficulty
     
    Last edited: May 2, 2006
  2. jcsd
  3. May 2, 2006 #2

    VietDao29

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    It's fine till here, but how can't you factor that?
    Do you know the formula to find the roots of a quadratic equation? Say, you want to solve for x in the equation:
    ax2 + bx + c = 0.
    [tex]x = \frac{-b \pm \sqrt{b ^ 2 - 4ac}}{2a}[/tex], does this formula look farmiliar to you? :)
    Can you go from here? :)
     
  4. May 2, 2006 #3
    I tried that one first. That also didn't work. That gave me (1+sqrt33)/8, which you would put within the arcsin function

    plugging it all in, it didn't work... instead of zero, i got 12 or something.
     
  5. May 3, 2006 #4

    VietDao29

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    [tex]\frac{1 + \sqrt{33}}{8}[/tex] is correct.
    So you'll have:
    [tex]\left[ \begin{array}{l} \sin x = \frac{1 + \sqrt{33}}{8} \\ \sin x = \frac{1 - \sqrt{33}}{8} \end{array} \right.[/tex]
    Can you find x? Note that [tex]x \in [0 , \ 2 \pi ][/tex] as the preblem states. There should be 4 x's. :)
     
    Last edited: May 3, 2006
  6. May 3, 2006 #5
    right, wouldn't it be

    [tex]\arcsin{x}=\frac{1+\sqrt(33)}{8}[/tex]

    and viceaversa for the 1-sqrt33


    ...however, when i plug that into the original equation, i don't get zero. this is what i'm plugging in:

    [tex]-(\frac{1+\sqrt(33)}{8}-2\cos(\arcsin(\frac{1+\sqrt(33)}{8})))[/tex]

    With an approximate value provided by the calculator, i get -1.9
     
  7. May 3, 2006 #6

    VietDao29

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    Ack, you've done it wrong...
    It should be:
    [tex]x = \arcsin \left( \frac{1 \pm \sqrt{33}}{8} \right)[/tex].
    You should also note that:
    [tex]\sin(x) = \sin (x + 2 \pi)[/tex]
    [tex]\sin(x) = \sin (\pi - x)[/tex].
    Can you find the 4 x's? :)
     
  8. May 3, 2006 #7
    I'm sorry... I still don't quite follow.

    You're right on the x=arcsin(...) bit, I wrote that incorrectly... But I plugged the correct bit into the above equation...

    I'm just a bit confused... It just looks like an impossible problem to solve, and I'm still stuck.
     
  9. May 3, 2006 #8
    I mean, I get your point, I get that it's

    [tex]x = \arcsin \left( \frac{1 + \sqrt{33}}{8} \right)[/tex]
    [tex]x = \arcsin \left( \frac{1 - \sqrt{33}}{8} \right)[/tex]

    and then the other two have something to do with
    [tex]x = \arcsin \left( \pi - \frac{1 + \sqrt{33}}{8} \right)[/tex]
    [tex]x = \arcsin \left( \pi - \frac{1 - \sqrt{33}}{8} \right)[/tex]


    But when I went and plugged the first one into the original equation, I didn't get 0...

    For cos[arcsin(...)]
    Isn't there some sort of trigonometric trick or something? Or do I have to somehow use and label the triangle...?
     
    Last edited: May 3, 2006
  10. May 3, 2006 #9

    Curious3141

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    Are you sure you're working in radians ? It won't make a difference in the first set of solutions, but it will make a difference in the second set (the ones with "pi" in them).

    Those two are right.

    No, these should read

    [tex]x = \pi - \arcsin \left(\frac{1 + \sqrt{33}}{8} \right)[/tex]
    [tex]x = \pi - \arcsin \left(\frac{1 - \sqrt{33}}{8} \right)[/tex]
     
    Last edited: May 3, 2006
  11. May 3, 2006 #10

    VietDao29

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    Uhmm, where did you get stuck?
    Okey, I'll give you an example:
    Find [tex]x \in [0 , \ \pi ][/tex], such that sin(x) = 0.4
    All the x that satisfy the above equation is:
    [tex]\left[ \begin{array}{l} x = \arcsin (0.4) + 2k \pi = 0.411 + 2k \pi \\ x = \pi - \arcsin (0.4) + 2m \pi = 2.73 + 2m \pi \end{array} \right. k, \ m \in \mathbb{Z}[/tex]
    Plug some value of k in to see if [tex]x \in [0 , \ \pi ][/tex].
    k = -1 ----> x = -5.8
    k = 0 ----> x = 0.41
    k = 1 ----> x = 6.7
    So k = 0 will give 1 valid x.
    Do the same for m, and you will have another valid x: x = 2.7 when m = 0.
    So the 2 solutions are:
    [tex]\left[ \begin{array}{l} x = \arcsin (0.4) \\ x = \pi - \arcsin (0.4) \end{array} \right.[/tex]
    Can you get this? :)
    ------------------
    Note that:
    [tex]x = \arcsin \left( \frac{1 - \sqrt{33}}{8} \right) < 0[/tex]
    By the way, what do you get if you plug that x in?
    Remember that it is:
    -sin x - 2cos(2x)
    Don't forget the factor of 2. :)
     
    Last edited: May 3, 2006
  12. May 3, 2006 #11
    You're right, I was missing the 2 infront of the x within the cosine function...

    However, now I get a really small number. The calculator is returning -1.9e-13

    I mean, it's close to zero... i mean, is it close enough?
     
  13. May 3, 2006 #12

    VietDao29

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    That's ok, it's just rounding error, don't worry about it.
    Have you found out the 4 x's? :)
     
  14. May 3, 2006 #13
    arcsin(1+...)
    arcsin(1-...)

    pi-arcsin(1+...)
    pi-arcsin(1-...)


    Correct?


    Thanks for all of your help!
     
  15. May 3, 2006 #14

    VietDao29

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    As I said earlier (in post # 10) that:
    [tex]x = \arcsin \left( \frac{1 - \sqrt{33}}{8} \right) < 0[/tex] is not a valid solution by the restriction [tex]x \in [ 0, \ 2 \pi ][/tex]
    Remember that sine is periodic function with period [tex]2 \pi[/tex], you can go from here? Right? :)
     
  16. May 3, 2006 #15
    well sure, I guess it would then be:

    pi-arcsin(1+...)
    arcsin(1+...)

    2pi+arcsin(1-...)
    pi-arcsin(1-...)

    would this be correct?
    all are greater than 0, i believe...
     
  17. May 3, 2006 #16

    VietDao29

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    That's correct. Congratulations. :)
     
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