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Few Trigonometric Functions that I can’t solve involving identities? helpp

  1. Mar 12, 2012 #1
    1. Sin^2(x) = 3 – x

    Answer: 2.97
    Attempts:
    1-cos^2(x) = 3 – x
    cos^2(x) - x + 2 = 0
    Factored it and got x = pi = 3.14
    It’s a multiple choice question, and other answers were 3.02,3.09 which are few decimal places off so the answer must not be pi since it's not even a choice. Is the answer key wrong or what?

    2. logx = 2 cos x, 0<x<2pi

    Answer: 1.38, 5.07
    Attempts:
    plug and chugged multiple choice answers, don't understand it

    2. Relevant equations

    sin2x = 2sinxcosx
    cos2x = 2cos^2(x) - 1
    cos2x = 1 - 2sin^2(x)
    1 = cos^2(x) + sin^2(x)
    some other trigonometric identities

    Thanks
     
  2. jcsd
  3. Mar 12, 2012 #2
    What class is this for? I'm not exactly sure if you can get an exact value for x in either equation, at least I didn't find any way to. Are you allowed to use Newton's Method to approximate the answer?
     
  4. Mar 12, 2012 #3

    NascentOxygen

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    With it being a multi choice question, wouldn't the most expedient course be to substitute the answer candidates one at a time? If it solves the equation, then it's a solution! http://img51.imageshack.us/img51/9206/t2710.gif [Broken]

    Or maybe you did this, and are now curious to know how to solve analytically?
     
    Last edited by a moderator: May 5, 2017
  5. Mar 12, 2012 #4

    NascentOxygen

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    π radians, you think? But sin(π) = 0, so you can see this can't be a solution. :frown:

    2.97127 is about right.
     
  6. Mar 12, 2012 #5

    Mark44

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    I have no idea how you were able to factor cos2(x) - x + 2 = 0 to come up with x = [itex]\pi[/itex].

    I thought you might have mistakenly used the Quadratic Formula (the equation is not a quadratic in x) on x2 - x + 2 = 0, but for this equation, the solutions are complex.
     
    Last edited: Mar 13, 2012
  7. Mar 12, 2012 #6
    This is for Math 12, which is the course right before Calculus 12.

    I factored it like
    (cosx-2)(cosx+1), but now I realize that would give cos^2(x) - x - 2 = 0,
    not cos^2(x) - x + 2 = 0

    I did plug and chug for the answer from the multiple choice, but I'm curious as to how to get the answer.
    Another way I could think of solving the question is by graphing it, but we aren't allowed to use a graphing calculator on the test and the values are so close...a graphing calculator wouldn't give the exact intersection anyways. y = Sin^2(x), y = 3 – x
     
  8. Mar 12, 2012 #7
    [itex](cos(x)-2)(cos(x)+1) = cos^{2}(x) - cos(x) - 2[/itex], not [itex]cos^{2}(x) - x - 2[/itex]

    I don't really see any way to isolate x.

    A nice way to solve this is to use Newton's Method to approximate the answer. Every iteration gets closer and closer to the answer provided your first guess was close enough. Have you gone over Newton's Method?
     
  9. Mar 12, 2012 #8
    I believe questions such as these where the answers are approximations generally involve plugging both sides of the equation as separate functions into your graphing calculator and finding the intersection, if this is indeed for Math 12. Have you tried that?
     
  10. Mar 13, 2012 #9

    NascentOxygen

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    That will work only where the alternative (and wrong) answers are adequately distanced from the correct answer. If there are two or more alternatives that are close to the graphical solution, the student will still need to evaluate the expressions to see which alternative is closest to solving the equation.
     
  11. Mar 13, 2012 #10
    Are you sure? Cause when I graph them I get 2.97 for question one, and two intersections for question two, 1.48 and 5.07.
     
  12. Mar 13, 2012 #11

    NascentOxygen

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    That's certainly true. http://fooplot.com/index.php?&type0=0&type1=0&type2=0&type3=0&type4=0&y0=2*cos%28x%29&y1=log%28x%29&y2=&y3=&y4=&r0=&r1=&r2=&r3=&r4=&px0=&px1=&px2=&px3=&px4=&py0=&py1=&py2=&py3=&py4=&smin0=0&smin1=0&smin2=0&smin3=0&smin4=0&smax0=2pi&smax1=2pi&smax2=2pi&smax3=2pi&smax4=2pi&thetamin0=0&thetamin1=0&thetamin2=0&thetamin3=0&thetamin4=0&thetamax0=2pi&thetamax1=2pi&thetamax2=2pi&thetamax3=2pi&thetamax4=2pi&ipw=0&ixmin=-5&ixmax=5&iymin=-3&iymax=3&igx=1&igy=1&igl=1&igs=0&iax=1&ila=1&xmin=-0.46432987052446434&xmax=6.611849756856397&ymin=-1.9401417847213813&ymax=1.519613137471211 [Broken]

    But the test question was probably worded "which is nearest to the correct answer."
     
    Last edited by a moderator: May 5, 2017
  13. Mar 13, 2012 #12

    SammyS

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    I agree with tal444 on the first --- close on the other..

    2.971269450098621372...


    1.401289...

    5.78291755...
     
  14. Mar 13, 2012 #13
    I'm just offering this as an alternate suggestion to i_m_mimi.

    Using Newton's Method for number two you can estimate the answer in your calculator by:

    [itex]Value \approx Ans - \displaystyle\frac{ln(Ans)-2cos(Ans)}{\frac{1}{Ans}+2sin(Ans)}[/itex]

    Most calculators can repeat the previous iteration of a formula they input. If you take the answer from the previous iteration (Ans in the formula above) and input it in the formula, you can create a recursive formula.

    If your graphing calculator can't pinpoint exactly what the intersection is or to not enough decimal points, this method can give you as many decimal points as your calculator will allow.

    Trying the formula above for starting values of 2 and 5, I got 5.78291755 radians and 1.401289368 radians for the two answers.
     
  15. Mar 14, 2012 #14
    if this is for a class called math 12 which is a prereq for calc 12, Newton's Method is probably not what they were looking for and will not mean much to i_m_mimi until next semester. My guess is that the teacher

    a) wants you to learn to make reasonable guesses when there is not analytic solution
    b) is lazy and doesn't want to come up with meaningful questions
     
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