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Math Help - Solving trigonometric equations?

  1. #1
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    Solving trigonometric equations?

    Hi, I'm working on this problem, and it's throwing me off a bit because it's a little different from what I've been doing.

    Solve 4(cos^2(3x)) 3 = 0 for [0, 2π). Give exact answers.

    Here is what I've done so far:

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

    4 (cos^2(3x)) = 3 (added 3)

    cos^2(3x) = 3/4 (divided by 4)


    The three is throwing me off a little and I don't know what the next step is.

    Please let me know if I can give anymore information or make anything clear.
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  2. #2
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    Re: Solving trigonometric equations?

    Quote Originally Posted by FatimaA View Post
    Hi, I'm working on this problem, and it's throwing me off a bit because it's a little different from what I've been doing.

    Solve 4(cos^2(3x)) 3 = 0 for [0, 2π). Give exact answers.

    Here is what I've done so far:

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

    4 (cos^2(3x)) = 3 (added 3)

    cos^2(3x) = 3/4 (divided by 4)


    The three is throwing me off a little and I don't know what the next step is.

    Please let me know if I can give anymore information or make anything clear.
    You have cos^2(3x) = 3/4. That square in the cosine is getting in the way, don't you think?

    -Dan
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  3. #3
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    Re: Solving trigonometric equations?

    Thank you, Dan.

    I determined:

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

    4 (cos^2(3x)) = 3 (added 3)

    cos^2(3x) = 3/4 (divided by 4)

    √cos^2(3x) = √3/4

    cos3x = +/- √3/(2)

    3x = π/6 + (nπ)

    and

    3x = 5π/6 + nπ

    Final answer (dividing by 3)

    x = π/18 + (nπ)/3

    and

    x = (5π)/18 + (nπ)/3
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  4. #4
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    Re: Solving trigonometric equations?

    Since \displaystyle \begin{align*} \cos{(3x)} = \pm \frac{\sqrt{3}}{2} \end{align*}, you should find that \displaystyle \begin{align*} 3x \end{align*} can give FOUR solutions (one for each quadrant of the unit circle), and then dividing by 3 will give you 12 solutions (as the period is diminished to \displaystyle \begin{align*} \frac{1}{3} \end{align*} of the original period).
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    Re: Solving trigonometric equations?

    Hi Prove it,

    I determined that there were the four solutions and then, dividing by three, I came up with:

    x = π/18 + (nπ)/3

    and

    x = (5π)/18 + (nπ)/3


    as my final answers.
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  6. #6
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    Re: Solving trigonometric equations?

    You have only gotten TWO of the possible solutions from the unit circle. There are FOUR.
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  7. #7
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    Re: Solving trigonometric equations?

    x = π/18 + (nπ)/3

    and

    x = (5π)/18 + (nπ)/3

    and

    x = (7π)/18 + (nπ)/3

    and

    x = (11π)/18 + (nπ)/3
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  8. #8
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    Re: Solving trigonometric equations?

    That's better, although you should be adding \displaystyle \begin{align*} \frac{2n\pi}{3} \end{align*} to each, not \displaystyle \begin{align*} \frac{n\pi}{3} \end{align*}.

    Now that you have the four starting solutions, what are the twelve solutions in the region \displaystyle \begin{align*} x \in [0 , 2\pi ) \end{align*}?
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