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4 Quick Math Questions for Grade 11

  1. May 16, 2005 #1




    [tex]This~is~not~a~quick~question~but~I~think~it~is~tough.~I~got~1~out~of~9~in~it.[/tex] :grumpy:

    [tex]6Cosx^2x+Sinx-4=0~for~ 0^0<x<360^0[/tex]

    [tex]4)~How~many~solutions~does~the~equation~\sqrt 2cos(2x)=1~for~0<x<2pi~have?[/tex]


    [tex]THANK~YOU[/tex] :smile:
    Last edited: May 16, 2005
  2. jcsd
  3. May 16, 2005 #2
    You give it a try first.
  4. May 16, 2005 #3
    I know the answer to the 1st question is -0.86602540378443864676372317075294 but how do I convert this into a fraction? My guess the answer has to be [tex]Cos(\frac{5pi}{1})[/tex]

    As for the 2nd question, my teacher gave me a zero on this question and I don't know what I did wrong.

    For the 3rd question,
    [tex]6Cosx^2x+Sinx-4=0~for~ 0^0~<0<360^0[/tex] factors into [tex](3Cosx-2)(3Cosx+2)[/tex]
    | |
    | |
    | |
    | |-------------->[tex]2Cosx+2=0[/tex]
    | |-------------->[tex]Cosx=-\frac{2}{2}[/tex]
    | |-------------->[tex]Cosx=-1[/tex]
    Last edited: May 16, 2005
  5. May 16, 2005 #4
    -0.86602540378443864676372317075294 reduces to a very common ratio in trig. Do you have a unit circle handy? Or some trig tables? Maybe drawing out a 30-60-90 triangle with hypotenuse 1, short side 0.5. Solve for the long side.
  6. May 16, 2005 #5
    Thank you for help me with question 1, it's [tex]-\frac{\sqrt3}{2}[/tex]
  7. May 16, 2005 #6

    James R

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    The factorisation question...

    [tex]\cos x - 2 \sin x \cos x[/tex]

    You answered:

    [tex](\cos x-\sin x)(\cos x+ \sin x)[/tex]

    But if we expand that out we get:

    [tex]\cos^2 x - \sin^2 x[/tex]

    which is obviously not the same as the original expression. The best you can do for the original expression is:

    [tex]\cos x - 2 \sin x \cos x = \cos x(1 - 2 \sin x)[/tex]
  8. May 16, 2005 #7
    If you are allowed double angles, you can get
    [tex] \cos x - \sin 2x [/tex] which is a bit simpler.
  9. May 16, 2005 #8
    I understand how to do 1 and 2 (Thank You) but how would you do the 4th one? That is the one where I got 1 out 9 on the unit test.
  10. May 16, 2005 #9
    [tex] \sqrt 2 \cos 2x = 1 [/tex] for x.

    Divide both sides by [itex] \sqrt 2 [/itex].

    Then for simplicity, say u = 2x.

    [tex] \cos u = \frac{1}{\sqrt{2}} = \frac{\sqrt 2}{2} [/tex]

    Solving that gives you u. Then use the u = 2x equation above to find x.
  11. May 16, 2005 #10
    I have tried to figure out question 3 myself. Could someone please check my work?

    [tex]6Cos^2x+Sinx-4=0~for~ 0^0[/tex]



    [tex]1-Sin^2x+Sinx=\frac{4}{6}[/tex] *Rearrange it as Sin2x+1-Sinx and factor it


    For (Sinx-1)

    For (Sinx+1)
  12. May 16, 2005 #11
    Between lines 3 and 4, you divide both sides by 6, you must also divide the single (sinx) term by 6.

    Also, between lines 4 and 5 you factor [itex] 1 - sin^2x + sinx [/itex] incorrectly. You seem to have trouble with factoring, try reviewing some. When you expand the binomials you wrote you get [itex] sin^2x - 1 [/tex] which isnt equal to the original expression.

    Also note, sinx is always bounded by -1 and 1. Your two answers lie outside these bounds.
  13. May 17, 2005 #12


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    I see a problem. On the left side you neglected to divide [itex] \frac{Sinx}{6} [/itex]
    then you would have: [itex] 1-Sin^2x+\frac{Sinx}{6}=\frac{4}{6} [/itex] which isn't too easy to work with.

    instead, how about multiplying through and factoring?
    [itex]-6Sin^2x+Sinx+2=0[/itex] .. you get the idea? :rolleyes:

    [edition]: I see whozum spotted the same trouble.. If you factor as I suggest, you will get two solutions that do lie on the unit circle (bounded by +1 and -1)
    Last edited: May 17, 2005
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