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Thread: Trigo graph stucked @ b

  1. January 19th 2010, 03:34 AM #1
    Punch
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    Trigo graph stucked @ b

    a) Sketch the graph of y=|2cosx|+1 for 0 ≤ x ≤ 2 \pi, stating the coordinates of the turning points.

    b) State the number of solutions for the equation |2cosx|+1 = \frac{x}{2} for 0 ≤ x ≤ 2 \pi
    I have already solved for part a, having problems with part b...

    I am guessing that we have to equate y=\frac{x}{2}
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  2. January 19th 2010, 04:33 AM #2
    Prove It
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    Quote Originally Posted by Punch View Post
    a) Sketch the graph of y=|2cosx|+1 for 0 ≤ x ≤ 2 \pi, stating the coordinates of the turning points.

    b) State the number of solutions for the equation |2cosx|+1 = \frac{x}{2} for 0 ≤ x ≤ 2 \pi
    I have already solved for part a, having problems with part b...
    Have you considered graphing the functions over that domain?
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  3. January 19th 2010, 05:50 AM #3
    Punch
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    Quote Originally Posted by Prove It View Post
    Have you considered graphing the functions over that domain?
    Yes, drawing a line would solve the question. However, I do not know how to draw the line y=\frac{x}{2} on this graph...
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  4. January 19th 2010, 06:01 AM #4
    mathaddict
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    Quote Originally Posted by Punch View Post
    Yes, drawing a line would solve the question. However, I do not know how to draw the line y=\frac{x}{2} on this graph...
    when x=2 , y=1

    when x=4 , y=2

    now you have 2 points , connect the 2 points and wherever it intersects your first function would be the solutions .
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  5. January 19th 2010, 06:03 AM #5
    Punch
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    Quote Originally Posted by mathaddict View Post
    when x=2 , y=1

    when x=4 , y=2

    now you have 2 points , connect the 2 points and wherever it intersects your first function would be the solutions .
    But on the x-axis is all in \pi... Do i convert \pi to values?
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  6. January 19th 2010, 06:19 AM #6
    mathaddict
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    Quote Originally Posted by Punch View Post
    But on the x-axis is all in \pi... Do i convert \pi to values?
    pi is 3.142 (in radians) .
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