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Math Help - Solve for value

  1. #1
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    Post Solve for value

    Find the exact values of x between 0 and 360 for

    2 cos^3(x) - cos(x) = 0

    -----------------------------------

    So what i did was cos(x) = x

    therefore 2x^3 - x = 0

    Now i got stuck there??
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  2. #2
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    Quote Originally Posted by taurus View Post
    Find the exact values of x between 0 and 360 for

    2 cos^3(x) - cos(x) = 0

    -----------------------------------

    So what i did was cos(x) = x

    therefore 2x^3 - x = 0

    Now i got stuck there??
    \cos x(2\cos^2 x - 1)=0
    Use identity,
    \cos x \cos 2x = 0
    Thus,
    \cos x = 0 \mbox{ or }\cos 2x = 0.
    Now continue ...
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  3. #3
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    ?

    Quote Originally Posted by ThePerfectHacker View Post
    \cos x(2\cos^2 x - 1)=0
    Use identity,
    \cos x \cos 2x = 0
    Thus,
    \cos x = 0 \mbox{ or }\cos 2x = 0.
    Now continue ...
    still a bit confused. so i use identity sin(2x) = cos(2x)?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by taurus View Post
    Find the exact values of x between 0 and 360 for

    2 cos^3(x) - cos(x) = 0
    The trig identity approach is confusing you. And please refer again to ThePerfectHacker's post: it is not sin(2x) = cos(2x). This is not an identity! The identity he used is cos(2x) = 2cos^2(x) - 1.

    Let's try it this way:
    2cos^3(x) - cos(x) = 0 <-- Factor the common cos(x)

    cos(x)(2cos^2(x) - 1) = 0

    Now either
    cos(x) = 0 \implies x = 90^o, 270^o
    or
    2cos^2(x) - 1 = 0 \implies cos(x) = \pm \sqrt{ \frac{1}{2}} \implies x = 45^o, 135^o, 225^o, 315^o

    -Dan
    Last edited by topsquark; September 4th 2007 at 11:33 AM.
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    Quote Originally Posted by topsquark View Post
    The trig identity approach is confusing you. And please refer again to ThePerfectHacker's post: it is not sin(2x) = cos(2x). This is not an identity! The identity he used is cos(2x) = 2cos^2(x) - 1.

    Let's try it this way:
    2cos^3(x) - cos(x) = 0 <-- Factor the common cos(x)

    cos(x)(2cos^2(x) - 1) = 0

    Now either
    cos(x) = 0 \implies x = 90^o, 180^o
    or
    2cos^2(x) - 1 = 0 \implies cos(x) = \pm \sqrt{ \frac{1}{2}} \implies x = 45^o, 135^o, 225^o, 315^o

    -Dan
    im still not quite getting it, ur saying cos(2x) but i dotn have cos(2x) in my expression? ummmm sorry
    you lost me by the factoring?

    also here:
    cos(x)(2cos^2(x) - 1) = 0

    whats happening there?

    thanks
    Last edited by taurus; September 4th 2007 at 05:59 AM.
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  6. #6
    Bar0n janvdl's Avatar
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    2cos^3 (x) - cos (x) = 0

    Set cos(x) = k

    2k^3 - k = 0

    k(2k^2 - 1) = 0

    Now set k = cos(x) like it was

    cos x ((2cos^2 x) - (1)) = 0

    But we know that 2cos^2 (x) - 1 = cos 2x from our identities

    So then (cos x)(cos 2x) = 0

    For Cos x = 0

    x = Cos^{-1} \ 0

    x = 90 \ or \ x = 270


    For Cos 2x = 0

    2x = Cos^{-1} \ 0

    x = \frac{Cos^{-1} \ 0}{2}

    x = 45 \ or \ x = 135 \ or \ x = 225 \ or \ x = 315
    Last edited by janvdl; September 4th 2007 at 07:29 AM.
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  7. #7
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    thanks

    ok right, so u jus solve for each of those cos
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  8. #8
    Bar0n janvdl's Avatar
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    Quote Originally Posted by taurus View Post
    ok right, so u jus solve for each of those cos
    Yes

    Oh and do refrain from using chat language, it's against the rules and you'll get in trouble.
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  9. #9
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    wait ur answers r different to that other guys?
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  10. #10
    Bar0n janvdl's Avatar
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    Quote Originally Posted by taurus View Post
    wait ur answers r different to that other guys?
    He made a mistake saying cos(180) = 0
    I'll check his others now.

    EDIT: His other solutions are correct.

    EDIT(2): Sorry i only forgot to write 225 and 315 as solutions. You will notice that  Cos(2x) = 0 is equal to 4 answers.
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by janvdl View Post
    He made a mistake saying cos(180) = 0
    Whoopsie! I fixed it, thanks for the spot.

    -Dan
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  12. #12
    Bar0n janvdl's Avatar
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    Quote Originally Posted by topsquark View Post
    Whoopsie! I fixed it, thanks for the spot.

    -Dan
    You do realise Topsquark that I have printed that one out and framed it against my wall...
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by janvdl View Post
    You do realise Topsquark that I have printed that one out and framed it against my wall...
    Well because it was you that corrected me I seriously thought about coming up with a really complicated argument involving Green's functions and doing a lot of chest thumping, but I decided not to.

    -Dan
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