Results 1 to 8 of 8

Math Help - Trig Equation

  1. #1
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    Trig Equation

    How do I solve 6cos(3x) = 3 without using the unit circle?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    6\cos(3x) = 3

    \cos(3x) = \frac{1}{2}

    3x = \frac{\pi}{3}

    x = \frac{\pi}{9}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    May 2009
    Posts
    527
    Are there any restrictions on x? Usually we solve with the restriction 0 \le x < 2\pi. If that's the case, then we'll need to expand on pickslides' work a bit. If we have the restriction 0 \le x < 2\pi, then it follows that 0 \le {\color{red}3}x < {\color{red}6}\pi.

    Quote Originally Posted by pickslides View Post
    6\cos(3x) = 3

    \cos(3x) = \frac{1}{2}

    3x = \frac{\pi}{3}
    Between 0 and 2 pi we also have 5pi/3 as a possible answer:

    3x = \frac{5\pi}{3}

    For each answer, we need to find coterminal angles to it that is in 0 \le 3x < 6\pi, so we add 2pi and 4pi to each:

    3x = \frac{\pi}{3}
    3x = \frac{\pi}{3} + 2\pi = \frac{7\pi}{3}
    3x = \frac{\pi}{3} + 4\pi = \frac{13\pi}{3}
    3x = \frac{5\pi}{3}
    3x = \frac{5\pi}{3} + 2\pi = \frac{11\pi}{3}
    3x = \frac{5\pi}{3} + 4\pi = \frac{17\pi}{3}

    Now take each equation and divide both sides by 3 to solve for x. So we'll get
    x = \frac{\pi}{9},\;\frac{5\pi}{9},\;\frac{7\pi}{9},\;  \frac{11\pi}{9},\;\frac{13\pi}{9},\;\frac{17\pi}{9  },\;

    These are all possible values of x that is in 0 \le x < 2\pi.


    01
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Yeongil is correct to show 6 solutions, although given the following

    Quote Originally Posted by magentarita View Post
    without using the unit circle?
    I would only give the 1 solution using the triangle made with angles \frac{\pi}{2},\frac{\pi}{3} and \frac{\pi}{6} ignoring the unit circle.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    more...

    Quote Originally Posted by pickslides View Post
    Yeongil is correct to show 6 solutions, although given the following



    I would only give the 1 solution using the triangle made with angles \frac{\pi}{2},\frac{\pi}{3} and \frac{\pi}{6} ignoring the unit circle.
    I need more than one or two answers.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    Exactly...

    Quote Originally Posted by yeongil View Post
    Are there any restrictions on x? Usually we solve with the restriction 0 \le x < 2\pi. If that's the case, then we'll need to expand on pickslides' work a bit. If we have the restriction 0 \le x < 2\pi, then it follows that 0 \le {\color{red}3}x < {\color{red}6}\pi.



    Between 0 and 2 pi we also have 5pi/3 as a possible answer:

    3x = \frac{5\pi}{3}

    For each answer, we need to find coterminal angles to it that is in 0 \le 3x < 6\pi, so we add 2pi and 4pi to each:

    3x = \frac{\pi}{3}
    3x = \frac{\pi}{3} + 2\pi = \frac{7\pi}{3}
    3x = \frac{\pi}{3} + 4\pi = \frac{13\pi}{3}
    3x = \frac{5\pi}{3}
    3x = \frac{5\pi}{3} + 2\pi = \frac{11\pi}{3}
    3x = \frac{5\pi}{3} + 4\pi = \frac{17\pi}{3}

    Now take each equation and divide both sides by 3 to solve for x. So we'll get
    x = \frac{\pi}{9},\;\frac{5\pi}{9},\;\frac{7\pi}{9},\;  \frac{11\pi}{9},\;\frac{13\pi}{9},\;\frac{17\pi}{9  },\;

    These are all possible values of x that is in 0 \le x < 2\pi.


    01
    This is exactly what I needed.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    539
    Hello: Here are the details for the solution :

    The general solution is :

    I'have :
    but :
    Calculate :
    Case 1 :
    but :
    :

    Case 2 :

    We find :

    :

    Thank you for all
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    wow!

    Quote Originally Posted by dhiab View Post
    Hello: Here are the details for the solution :

    The general solution is :

    I'have :
    but :
    Calculate :
    Case 1 :
    but :
    :

    Case 2 :

    We find :

    :

    Thank you for all
    A very detailed reply!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compute Trig Function Values, Solve Trig Equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 8th 2011, 08:00 PM
  2. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 14th 2011, 08:07 AM
  3. Trig Equation with varied trig functions
    Posted in the Trigonometry Forum
    Replies: 12
    Last Post: April 12th 2010, 11:31 AM
  4. Replies: 1
    Last Post: July 24th 2009, 04:56 AM
  5. Replies: 1
    Last Post: July 24th 2009, 03:29 AM

Search Tags


/mathhelpforum @mathhelpforum