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Math Help - Inequalities

  1. #1
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    Inequalities

    I was asked to solve \frac{x^2-x-2}{x-1}<2 from which the solution is x<0 or 1<x<3

    The next part asks: Hence solve \frac{cos^22\theta-cos2\theta-2}{cos2\theta-1}<2, where  0 &\leq \theta &\leq 2\pi

    I tried doing: Replace x with cos2\theta

    cos2\theta<0 or 1<cos2\theta<3

    therefore, \theta<\frac{\pi}{4} or 0<\theta< but cos^{-1}3=error

    Can you show me the proper way to solve such an inequality? I totally forgot how to tackle this...
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  2. #2
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    Re: Inequalities

    If 0\leq\theta\leq 2\pi, then \cos 2\theta<0 is equivalent to \pi/4<\theta<3\pi/4. Draw the unit circle and see for which angles cosine is negative.

    Concerning 1<\cos2\theta<3, it helps to draw a graph of \cos2\theta and see which portion is located between the lines y = 1 and y = 3. Can \cos2\theta ever be greater than 1?
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  3. #3
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    Re: Inequalities

    Quote Originally Posted by emakarov View Post
    If 0\leq\theta\leq 2\pi, then \cos 2\theta<0 is equivalent to \pi/4<\theta<3\pi/4. Draw the unit circle and see for which angles cosine is negative.

    Concerning 1<\cos2\theta<3, it helps to draw a graph of \cos2\theta and see which portion is located between the lines y = 1 and y = 3. Can \cos2\theta ever be greater than 1?
    no
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  4. #4
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    Re: Inequalities

    That's right. I hope you have the answer to the original in equality.
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  5. #5
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    Re: Inequalities

    I drew the circle and it shows that the 2nd and 3rd quadrant are where cos\theta<0 but the question is cos2\theta, shouldn't there be some changes? why is it still the same?
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  6. #6
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    Re: Inequalities

    It is not the same. As you said, we have \cos\alpha<0 iff \pi/2<\alpha<3\pi/2. In particular, for \alpha=2\theta, \cos2\theta<0 iff \pi/2<2\theta<3\pi/2 iff \pi/4<\theta<3\pi/4.
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  7. #7
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    Re: Inequalities

    nothing lies between the line y=1 and y=3, do i reject this solution? but it seems that the model answer has another solution of \frac{5\pi}{4}<\theta<\frac{7\pi}{4} in addition to the one you mentioned
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  8. #8
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    Re: Inequalities

    Quote Originally Posted by Punch View Post
    nothing lies between the line y=1 and y=3, do i reject this solution?
    Yes, no value of \theta can contribute to the solution of 1<\cos2\theta.

    Quote Originally Posted by Punch View Post
    but it seems that the model answer has another solution of \frac{5\pi}{4}<\theta<\frac{7\pi}{4} in addition to the one you mentioned
    You are right. I missed that 5\pi/2<2\theta<7\pi/2 also implies \cos2\theta<0, and even though the interval (5\pi/2,7\pi/2) is outside of [0,2\pi], the corresponding interval for \theta: (5\pi/4,7\pi/4) is inside [0,2\pi].

    It is better to be precise and to select the solutions from [0,2\pi] in the end. So, \cos2\theta iff \pi/2+2\pi k<2\theta<3\pi/2+2\pi k, k\in\mathbb{Z}. This is equivalent to \pi/4+\pi k<\theta<3\pi/4+\pi k, k\in\mathbb{Z}. Of these, only intervals for k = 0 and k = 1 lie inside [0,2\pi].
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