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Math Help - A series of problems... (don't be afraid it's only maths questions)

  1. #1
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    A series of problems... (don't be afraid it's only maths questions)

    Ok

    Q1) Solve (x^3) - 7x + 6 = 0

    Q2) ∫▒〖sec4xtan4x dx〗
    if the above isn't readable, i meant find: ∫sec6xtan6x dx

    Thanks for any help xD
    Last edited by CaptainBlack; August 6th 2010 at 02:40 AM.
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  2. #2
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    f(x) = x^3 - 7x + 6.

    It's pretty easy to see that f(1) = 0, so x - 1 is a factor.

    Long dividing gives

    f(x) = (x-1)(x^2 + x - 6)

     = (x - 1)(x + 3)(x - 2).


    So if x^3 - 7x + 6 = 0

    (x - 1)(x + 3)(x - 2) = 0

    x - 1 = 0 or x + 3 = 0 or x - 2 = 0.

    So x = -3 or x = 1 or x = 2.
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  3. #3
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    Quote Originally Posted by MathsWhiz View Post
    Ok

    Q1) Solve (x^3) - 7x + 6 = 0
    Have you tried anything at all? For example, have you looked at what happens if x= 1?

    Q2) ∫▒〖sec4xtan4x dx〗
    if the above isn't readable, i meant find: ∫sec6xtan6x dx
    How did the "4" in the possibly un readable form become "6"?
    sec 6x= 1/cos(6x) and tan(6x)= sin(6x)/cos(6x) so this is sin(6x)/cos^2(6x). Let u= cos 6x.

    Q3) Solve: log(x) + log(x+1) = 1
    log(x(x+1)= 1

    Q4) It is given cos(θ) = 5/13 and that sin(θ) is negative. Find the exact value of tan(θ/2).
    sin^2(\theta)+ cos^2(\theta)= 1 so you can find sin(\theta) easily and then tan(\theta). Then use the trig identity [tex]tan(\theta)= \frac{2tan(\theta/2)}{1+ tan^2(\theta/2)}. (I actually used sin(2\theta)= 2sin(\theta)cos(\theta) and cos(2\theta)= cos^2(\theta)- sin^2(\theta) to get a formula for tan(2\theta) and then swapped \theta and 2\theta.

    Q5) The graphs of y -4x^2 and y = 6 - 2x intersect at x =1. Find the size of the acute angle between these curves at x = 1 correct to the nearest minute.
    Find the tangent lines at x= 1 and find the angle between them. You might want to use the trig identity tan(\theta+ \phi)= \frac{tan(\theta)+ tan(\phi)}{1+ tan(\theta)tan(\phi)}. Again, I used the sum formulas for sine and cosine to get that formula.

    i) Express 3cosx - √3sinx in the form R cos (x +a)
    cos(\theta+ \phi)= cos(\theta)cos(\phi)- sin(theta)sin(\phi). With \phi= x, what must \theta be?

    Thanks for any help xD
    And the "giggle" and "lipssealed" emoticons are to indicate that anyone giving you answers will be helping you cheat?

    In future, try yourself and show what you have tried.
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    \log(x) + \log(x + 1) = 1

    \log[x(x + 1)] = 1

    \log(x^2 + x) = 1

    x^2 + x = e

    x^2 + x - e = 0


    x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-e)}}{2(1)}

     = \frac{-1 \pm \sqrt{1 + 4e}}{2}


    So x = \frac{-1 - \sqrt{1 + 4e}}{2} or x = \frac{-1 + \sqrt{1 + 4e}}{2}.
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  5. #5
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    Captain Black, did you edit to remove problems? Why?
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    Quote Originally Posted by HallsofIvy View Post
    Have you tried anything at all? For example, have you looked at what happens if x= 1?


    How did the "4" in the possibly un readable form become "6"?
    sec 6x= 1/cos(6x) and tan(6x)= sin(6x)/cos(6x) so this is sin(6x)/cos^2(6x). Let u= cos 6x.


    log(x(x+1)= 1


    sin^2(\theta)+ cos^2(\theta)= 1 so you can find sin(\theta) easily and then tan(\theta). Then use the trig identity [tex]tan(\theta)= \frac{2tan(\theta/2)}{1+ tan^2(\theta/2)}. (I actually used sin(2\theta)= 2sin(\theta)cos(\theta) and cos(2\theta)= cos^2(\theta)- sin^2(\theta) to get a formula for tan(2\theta) and then swapped \theta and 2\theta.


    Find the tangent lines at x= 1 and find the angle between them. You might want to use the trig identity tan(\theta+ \phi)= \frac{tan(\theta)+ tan(\phi)}{1+ tan(\theta)tan(\phi)}. Again, I used the sum formulas for sine and cosine to get that formula.


    cos(\theta+ \phi)= cos(\theta)cos(\phi)- sin(theta)sin(\phi). With \phi= x, what must \theta be?


    And the "giggle" and "lipssealed" emoticons are to indicate that anyone giving you answers will be helping you cheat?

    In future, try yourself and show what you have tried.
    If I knew how to do them, if I knew where to start, I wouldn't be asking for help would I?
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  7. #7
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    Quote Originally Posted by MathsWhiz View Post
    If I knew how to do them, if I knew where to start, I wouldn't be asking for help would I?
    The point is that you shouldn't be asking for people to do all the work for you without even attempting to do these problems yourself. It can be considered cheating.
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  8. #8
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    Thread closed.
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    Captain Black, did you edit to remove problems? Why?
    The post was edited while you and others were replying. The reason for the edit is given in rule #8.

    Regarding this thread: Rule #11 is also relevant. As is rule #4. Rule #6 may also be relevant (hence the current closure of the thread).
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