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Math Help - Solving Within an Interval & Finding the Period

  1. #1
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    Solving Within an Interval & Finding the Period

    1) Find the solutions of in the interval

    I can determine that there are 4 solutions that need to be found, but I can only find 2 of them (). How do I find the others?

    2) Find the period of the graph

    I know how to find the period of functions containing either sin or cos, but not both. Can someone give me a crash course on how to do this?


    Thanks in advance.
    (Apologies for the choppy equation structure.)
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  2. #2
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    Re: Solving Within an Interval & Finding the Period

    2\sin 3x +\sqrt{2}=0
    \sin 3x=-\frac{\sqrt{2}}{2}

    Draw a unit trigonometric circle, find the value -\frac{\sqrt{2}}{2} on y-axis, and . Draw a line parallel to x-axis that runs trough -\frac{\sqrt{2}}{2} on the y-axis. The line intersects the unit circle in two distinct points - numbers that are at those two points on the trigonometric circle are the solutions to the given equation.

    All of the solutions can be found solving  3x=\frac{5\pi}{4}+k\cdot 2\pi and  3x=\frac{7\pi}{4}+k\cdot 2\pi, k\in \mathbf{Z}.
    Last edited by MathoMan; August 24th 2013 at 05:13 PM.
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    Re: Solving Within an Interval & Finding the Period

    1) Remember that sin(3x)=sin(3x+2\pi)=sin(3x+4\pi) When you find a solution to the equation you can add 2*pi to it again and again to get more solutions but you have an upper limit so the number of solutions will be limited.

    2) The period P will satisfy the equation f(x)=f(x+P)
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    Re: Solving Within an Interval & Finding the Period

    I realise that my suggestion for part 2) is very difficult if not impossible to solve.
    You cannot guarantee this will be the smallest value of p to satisfy the equation but the function will repeat if there is a value y so that 4sin(3\pi x)=4sin(3\pi(x+y)) and -3cos(2\pi x)=-3cos(2\pi(x+y))

    Remember that sin(x+2\pi n)=sin(x)
    same for cos
    y will have to be an integer
    Last edited by Shakarri; August 24th 2013 at 06:01 PM.
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    Re: Solving Within an Interval & Finding the Period

    Quote Originally Posted by MathoMan View Post

    All of the solutions can be found solving  3x=\frac{5\pi}{4}+k\cdot 2\pi and  3x=\frac{7\pi}{4}+k\cdot 2\pi, k\in \mathbf{Z}.
    I'm not quite sure I fully understand. What's the value of k?
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