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Math Help - Trigonometric Equations

  1. #1
    Senior Member
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    Trigonometric Equations

    When the problem asks to Find all solutions of the equation, I can solve the equation perfectly fine but i'm having trouble understanding when to use

    kpi or 2kpi

    For instance:

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

    cos(x) = Positive or Negative 1/2

    well here is where I get confused as to how the answer should be and why.

    Here is the way my teacher interpreted it for us: Since Cosine has a period of 2pi, we find all the radian measures where Cosine is 1/2 in the full circle or 2pi.

    So I find : pi / 3, 2pi / 3, 4pi / 3, 5pi / 3.

    This is the first part where I don't understand. According to the teacher, the answer should be all those plus 2kpi since Cosine has a period of 2pi.

    Well the answers in the back of the book is claiming the answer is

    (pi / 3) + kpi and (2pi/3) + kpi


    I'm confused
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  2. #2
    MHF Contributor
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    4cos^2(x) = 1
    2cos(x) = +,-1
    cos(x) = +,-1/2

    When cos(x) = 1/2.
    Cosine is positive in the 1st and 4th quadrants, so,
    x = pi/3 and [2pi -pi/3 = 5pi/3] in the interval [0,2pi) or in one revolution.
    In many revolutions, [1 revolution = 360 degrees = 2pi radians],
    x = pi/3 +2k*pi, and 5pi/3 +2k*pi ---because the period or cycle of cosine is 2pi.

    When cos(x) = -1/2.
    Cosine is negative in the 2nd and 3rd quadrants, so,
    x = [pi -pi/3 = 2pi/3] and [pi +pi/3 = 4pi/3] in the interval [0,2pi) or in one revolution.
    In many revolutions,
    x = 2pi/3 +2k*pi, and 4pi/3 +2k*pi

    Hence,
    x = pi/3 +2k*pi, 2pi/3 +2k*pi, 4pi/3 +2k*pi, 5pi/3 +2k*pi
    Meaning, in all the 4 quadrants, x has a value or solution.
    Meaning, the solution repeats every 180 degrees or every pi.
    Meaning, the period or cycle here of cosine is 180deg or pi only. The effective period is no longer 360deg or 2pi.
    Hence, x = pi/3 +k*pi and 2pi/3 +k*pi.
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  3. #3
    Senior Member DivideBy0's Avatar
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    That notation is just a nice way of summarising the solutions, since

    \frac{4\pi}{3}=\frac{\pi}{3} + \pi

    and

    \frac{5\pi}{3}=\frac{2\pi}{3}+\pi

    One angle will have a period of 2\pi before it returns to itself, but here we have many angles, and it just happens that one angle is \pi greater than another angle, so the notation is more convenient.

    You could, I guess, say:

    x=\frac{\pi}{3}+2k\pi,\frac{2\pi}{3}+2k\pi,\frac{4  \pi}{3}+2k\pi,\frac{5\pi}{3}+2k\pi

    But what's the point?
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  4. #4
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    MmMm..... that way is confusing me a bit more haha. Is the method my teacher taught us wrong then?

    Cos(x) = +,- 1/2 which appears in all 4 quadrants.

    If the question is asking me to find all solutions

    Then essentially,

    (pi / 3) + 2kpi, (2pi / 3) + 2kpi, (4pi / 3) + 2kpi, and (5pi / 3) + 2kpi

    is correct?







    Quote Originally Posted by ticbol View Post
    4cos^2(x) = 1
    2cos(x) = +,-1
    cos(x) = +,-1/2

    When cos(x) = 1/2.
    Cosine is positive in the 1st and 4th quadrants, so,
    x = pi/3 and [2pi -pi/3 = 5pi/3] in the interval [0,2pi) or in one revolution.
    In many revolutions, [1 revolution = 360 degrees = 2pi radians],
    x = pi/3 +2k*pi, and 5pi/3 +2k*pi ---because the period or cycle of cosine is 2pi.

    When cos(x) = -1/2.
    Cosine is negative in the 2nd and 3rd quadrants, so,
    x = [pi -pi/3 = 2pi/3] and [pi +pi/3 = 4pi/3] in the interval [0,2pi) or in one revolution.
    In many revolutions,
    x = 2pi/3 +2k*pi, and 4pi/3 +2k*pi

    Hence,
    x = pi/3 +2k*pi, 2pi/3 +2k*pi, 4pi/3 +2k*pi, 5pi/3 +2k*pi
    Meaning, in all the 4 quadrants, x has a value or solution.
    Meaning, the solution repeats every 180 degrees or every pi.
    Meaning, the period or cycle here of cosine is 180deg or pi only. The effective period is no longer 360deg or 2pi.
    Hence, x = pi/3 +k*pi and 2pi/3 +k*pi.
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  5. #5
    MHF Contributor
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    Yes, of course. The method your teacher taught you is correct.

    The pi/3 +k*pi, 2pi/3 +k*pi just summarized your long answer.
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