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Math Help - inequalities

  1. #1
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    inequalities

    By means of graph, solve the inequality
    tan2x+1≤4

    Iíve tried like (2tanx)/(1-tan2) -3≤0
    then (3tan2x+2tanx-3)/(1+tanx)(1-tanx) ≤0
    Ö

    can anyone point out my mistake?
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  2. #2
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    Re: inequalities

    Have you actually graphed it?
    inequalities-tan-2x-.jpg

    What does the graph tell you? (The point is that you are not going to find a closed form solution so you estimate the solutions via graphing.)

    -Dan
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  3. #3
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    Re: inequalities

    Is that tan(2x)+ 1 or tan(2x+ 1)?
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    Re: inequalities

    Quote Originally Posted by HallsofIvy View Post
    Is that tan(2x)+ 1 or tan(2x+ 1)?
    tan(2x) +1
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    Re: inequalities

    If you have properly described the question, you are to solve this BY MEANS OF A GRAPH.

    As topsquark pointed out, this requires sketching the pertinent graph and no analysis at all.

    To deal with this analytically rather than geometrically, I'd probably start with u = 2x which reduces the problem to $tan(u) \le 3.$

    Then I'd use the inverse trig functions.
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    Re: inequalities

    Quote Originally Posted by JeffM View Post
    If you have properly described the question, you are to solve this BY MEANS OF A GRAPH.

    As topsquark pointed out, this requires sketching the pertinent graph and no analysis at all.

    To deal with this analytically rather than geometrically, I'd probably start with u = 2x which reduces the problem to $tan(u) \le 3.$

    Then I'd use the inverse trig functions.
    I understand that. And I've sketched the graph, but I'm not sure still..

    let tan2x=3
    2x=71.57, 251.57, 431.57, 611.57,
    x=35.79, 125.79, 215.79, 305.79
    so, 0≤x≤35.79, 45<x≤125.79, 135<x≤215.79, 225<x≤305.79, 315<x≤360??
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    Re: inequalities

    Quote Originally Posted by Trefoil2727 View Post
    I understand that. And I've sketched the graph, but I'm not sure still..

    let tan2x=3
    2x=71.57, 251.57, 431.57, 611.57,
    x=35.79, 125.79, 215.79, 305.79
    so, 0≤x≤35.79, 45<x≤125.79, 135<x≤215.79, 225<x≤305.79, 315<x≤360??
    How you answer depends on exactly how the question is asked and what you have been taught.

    Are you to answer in radians or degrees? Are there any additional limits placed on x? Do you know how to represent a cyclic answer?

    A cyclic answer in radians looks like this:

    $cos( \theta ) = 0 \implies \theta = \dfrac{\pi }{2} * (2k - 1)\ for\ all\ k \in \mathbb Z.$
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    Re: inequalities

    [QUOTE=JeffM;823365]How you answer depends on exactly how the question is asked and what you have been taught.

    Are you to answer in radians or degrees? Are there any additional limits placed on x? Do you know how to represent a cyclic answer?

    answer in degree. And there's no additional limits
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  9. #9
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    Re: inequalities

    $tan(2x) + 1 \le 4 \implies tan(2x) \le 3.$

    $0 < v < 90^o\ and\ v = arctan(3) \implies v \approx 71.5650^o < 90^o.$

    $-90^o < 2x < 90^o\ and\ tan(2x) \le 3 \implies -45^o < x\ and\ 2x \le v \implies - 45^o < x \le \dfrac{arcsin(3)}{2} \approx 35.7825^o.$

    So one possible answer is $- 45^o < x \le \dfrac{arcsin(3)}{2} \approx 35.7825^o.$

    A better answer might be $\{-45 + (2k - 1) * 90\}^o < x \le \left\{\dfrac{arcsin(3)}{2} + (2k - 1) * 90\right\}^o,\ for\ all\ k \in \mathbb Z.$
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  10. #10
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    Re: inequalities

    Quote Originally Posted by JeffM View Post
    $tan(2x) + 1 \le 4 \implies tan(2x) \le 3.$

    $0 < v < 90^o\ and\ v = arctan(3) \implies v \approx 71.5650^o < 90^o.$

    $-90^o < 2x < 90^o\ and\ tan(2x) \le 3 \implies -45^o < x\ and\ 2x \le v \implies - 45^o < x \le \dfrac{arcsin(3)}{2} \approx 35.7825^o.$

    So one possible answer is $- 45^o < x \le \dfrac{arcsin(3)}{2} \approx 35.7825^o.$

    A better answer might be $\{-45 + (2k - 1) * 90\}^o < x \le \left\{\dfrac{arcsin(3)}{2} + (2k - 1) * 90\right\}^o,\ for\ all\ k \in \mathbb Z.$
    Why are you using 2k-1? There is no possible value of k such that you get the inequality you gave (second line from bottom of your quote).
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  11. #11
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    Re: inequalities

    Ooops. Should have been 2k, not 2k - 1. No clue what cog slipped there.
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