Results 1 to 8 of 8

Thread: Inequalities

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Inequalities

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

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

    I tried doing: Replace $\displaystyle x$ with $\displaystyle cos2\theta$

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

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

    Can you show me the proper way to solve such an inequality? I totally forgot how to tackle this...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790

    Re: Inequalities

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

    Concerning $\displaystyle 1<\cos2\theta<3$, it helps to draw a graph of $\displaystyle \cos2\theta$ and see which portion is located between the lines y = 1 and y = 3. Can $\displaystyle \cos2\theta$ ever be greater than 1?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Re: Inequalities

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

    Concerning $\displaystyle 1<\cos2\theta<3$, it helps to draw a graph of $\displaystyle \cos2\theta$ and see which portion is located between the lines y = 1 and y = 3. Can $\displaystyle \cos2\theta$ ever be greater than 1?
    no
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790

    Re: Inequalities

    That's right. I hope you have the answer to the original in equality.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Re: Inequalities

    I drew the circle and it shows that the 2nd and 3rd quadrant are where $\displaystyle cos\theta<0$ but the question is $\displaystyle cos2\theta$, shouldn't there be some changes? why is it still the same?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790

    Re: Inequalities

    It is not the same. As you said, we have $\displaystyle \cos\alpha<0$ iff $\displaystyle \pi/2<\alpha<3\pi/2$. In particular, for $\displaystyle \alpha=2\theta$, $\displaystyle \cos2\theta<0$ iff $\displaystyle \pi/2<2\theta<3\pi/2$ iff $\displaystyle \pi/4<\theta<3\pi/4$.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Dec 2009
    Posts
    755

    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 $\displaystyle \frac{5\pi}{4}<\theta<\frac{7\pi}{4}$ in addition to the one you mentioned
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790

    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 $\displaystyle \theta$ can contribute to the solution of $\displaystyle 1<\cos2\theta$.

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

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

Similar Math Help Forum Discussions

  1. inequalities
    Posted in the Algebra Forum
    Replies: 7
    Last Post: Aug 20th 2011, 04:56 AM
  2. inequalities
    Posted in the Algebra Forum
    Replies: 0
    Last Post: Mar 20th 2011, 02:42 AM
  3. Inequalities
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Jul 1st 2009, 04:35 AM
  4. Inequalities
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Jun 15th 2009, 11:26 AM
  5. inequalities
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Oct 14th 2008, 09:57 PM

Search Tags


/mathhelpforum @mathhelpforum