Results 1 to 6 of 6

Math Help - help! adv math homework help!!!! plz?

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    3

    help! adv math homework help!!!! plz?

    square root of 3 times the cot times 3x plus one equals 0..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Question

    Quote Originally Posted by ericchung912 View Post
    square root of 3 times the cot times 3x plus one equals 0..
    The cotangent of what? Do you mean \sqrt3\cot\left(3x + 1\right) = 0? Or maybe \sqrt{3\cot\left(3x + 1\right)} = 0? Or \sqrt3\cot\left(3x\right) + 1 = 0? \sqrt{3\cot\left(3x\right)} + 1 = 0? \sqrt{3\cot\left(3x\right) + 1} = 0? \sqrt3\left[\cot\left(3x\right) + 1\right] = 0?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by ericchung912 View Post
    square root of 3 times the cot times 3x plus one equals 0..
    Is this what you mean?

    \sqrt{3}\cot(3x)+1=0

    Is there a restriction on x? (i.e. 0 \leq x <2\pi)?

    After some modifications, we get:

    \cot(3x)=-\frac{1}{\sqrt{3}}

    Note that \cot\theta=-\frac{1}{\sqrt{3}} when \theta=\frac{2\pi}{3}\text{,} \ \frac{5\pi}{3}\text{,} \ \dots \ \frac{(3n-1)\pi}{3}; \ n\in\mathbb{N}

    Thus, 3x=\frac{2\pi}{3}\text{,} \ \frac{5\pi}{3}\text{,} \ \dots \ \frac{(3n-1)\pi}{3}; \ n\in\mathbb{N}
    \implies x=\frac{2\pi}{9}\text{,} \ \frac{5\pi}{9}\text{,} \ \dots \ \frac{(3n-1)\pi}{3}; \ n\in\mathbb{N}.

    I hope this helped (assuming that I interpreted the question properly)!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2008
    Posts
    3
    thats what i mean ypu.. knda new. here so i dont know how to use it
    <br /> <br />
\sqrt{3}\cot(3x)+1=0<br />

    the directions to this problem is:
    solve the following equations that o is less than or equal to x...x is less than 360
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by ericchung912 View Post
    thats what i mean ypu.. knda new. here so i dont know how to use it
    <br /> <br />
\sqrt{3}\cot(3x)+1=0<br />

    the directions to this problem is:
    solve the following equations that o is less than or equal to x...x is less than 360
    Ok then! Referring back to my solution (I had it in radians, now I'm gonna convert it to degrees...), we saw that:

    x=\frac{2(360^o)}{9}\text{,} \ \frac{5(360^o)}{9}\text{,} \ \frac{8(360^o)}{9}\text{,} \ \frac{11(360^o)}{9}\text{,} \ \frac{14(360^o)}{9}\text{,} \ \dots \ \frac{(3n-1)(360^o)}{9}; \ n\in\mathbb{N}

    After simplifying we get:

    \color{red}\boxed{x=80^o\text{,} \ 200^o\text{,} \ 320^o}

    The ones after 320 are outside of the domain ( 0\leq x<360^o).

    Hope this made sense!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Chris L T521 View Post
    Ok then! Referring back to my solution (I had it in radians, now I'm gonna convert it to degrees...), we saw that:

    x=\frac{2(360^o)}{9}\text{,} \ \frac{5(360^o)}{9}\text{,} \ \frac{8(360^o)}{9}\text{,} \ \frac{11(360^o)}{9}\text{,} \ \frac{14(360^o)}{9}\text{,} \ \dots \ \frac{(3n-1)(360^o)}{9}; \ n\in\mathbb{N}

    After simplifying we get:

    \color{red}\boxed{x=80^o\text{,} \ 200^o\text{,} \ 320^o}

    The ones after 320 are outside of the domain ( 0\leq x<360^o).

    Hope this made sense!
    Forgive me, but I made a "little" error

    \pi = 180^o\neq 360^o

    Thus, \color{red}\boxed{x=40^o\text{,} \ 100^o\text{,} \ 160^o\text{,} \ 220^o\text{,} \ 280^o\text{,} \ 340^o}

    To get to the new answers, divide the old ones by 2 (you will have more answers...I hope you can follow).

    Sorry about that!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need Math Homework help?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: January 13th 2011, 09:19 PM
  2. Math Homework Help
    Posted in the Algebra Forum
    Replies: 7
    Last Post: September 30th 2010, 02:09 PM
  3. Help with math homework
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 22nd 2009, 02:14 PM
  4. Math Homework Help!
    Posted in the Algebra Forum
    Replies: 6
    Last Post: August 21st 2008, 10:38 AM
  5. [SOLVED] math homework
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 12th 2006, 05:02 PM

Search Tags


/mathhelpforum @mathhelpforum