Results 1 to 6 of 6

Math Help - Solving trig equation

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    39

    Solving trig equation

    Find solutions:

    sin ( \frac{1}{2} - 7x) \times \pi = cos(2-5x) \times \pi + sin8 \pi

    Interval - [ \frac{1}{2};1]
    -----------
    What I did: 1) sin ( \frac{1}{2} - 7x) = sin \frac{1}{2}cos7x + sin(7x)cos( \frac{1}{2})

    2) cos(2-5x) = cos2cos5x + sin2sin5x

    ..and 3) sin8 \pi = sin2 \pi

    But what should I do now, should I multiply by \pi ? Thank you.
    Last edited by Ellla; June 4th 2011 at 12:35 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Do you mean \sin \left(\dfrac{1}{2}-7x\right) \times \pi or \sin \left[\left \pi (\dfrac{1}{2}-7x\right)\right]

    Same for the other terms.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    39
    Quote Originally Posted by e^(i*pi) View Post
    Do you mean \sin \left(\dfrac{1}{2}-7x\right) \times \pi or \sin \left[\left \pi (\dfrac{1}{2}-7x\right)\right]

    Same for the other terms.
    Sorry, it is \sin \left(\dfrac{1}{2}-7x\right) \times \pi
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2011
    From
    Bjelovar, Croatia
    Posts
    16
    Thanks
    1
    Hi!

    So, your original equation was, if I'm reading it correctly:

     \pi\cdot\sin{\left(\frac{1}{2} - 7x\right)} = \pi\cdot\cos{\left(2-5x\right)} + \sin{8\pi}

    Now, considering that  \sin{8\pi}=0 and that the 'pi'-es on both sides cancel each other out, the equation turns to:

     \sin{\left(\frac{1}{2} - 7x\right)} = \cos{\left(\2-5x\right)}

    By squaring both sides, we'll be able to use an identity that'll get rid of the unsightly fraction in the sine:

     \sin^2{\left(\frac{1}{2} - 7x\right)} = \cos^2{\left(2-5x\right)}
     \frac{1 - \cos{(1-14x)}}{2} = \frac{1 + \cos{(4-10x)}}{2}
      -\cos{(1-14x)} = \cos{(4-10x)}
     -\cos{(1-14x)}-\cos{(4-10x)} = 0

    And after multiplying it by (-1), we convert it from a sum to a product:

     2\cos{\left(\frac{1-14x+4-10x}{2}\right)}\cos{\left(\frac{1-14x-4x+10x}{2}\right)} = 0

    Which is equivalent to:

     \cos{\left(\frac{5}{2} - 12x\right)}\cos{\left(-\frac{3}{2} - 2x\right)} = 0 ,

    and by equating each factor with zero, you get the solutions:

     x_1 = \frac{5-\pi}{24} + k\cdot\frac{\pi}{6}, \, k\in\mathbf{Z}
     x_2 = \frac{-3-\pi}{4} + k\cdot\pi, \, k\in\mathbf{Z}

    We find the solution within the interval \left[\frac{1}{2}, 1\right] by including arbitrary values of the integer 'k'. By including k=1, we find the only solution, derived from the first general solution, and that is:

     x = \frac{5+3\pi}{24},\, k=1

    Hope I helped!
    Last edited by mrfour44; June 5th 2011 at 04:34 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Quote Originally Posted by mrfour44 View Post
     \sin{\left(\frac{1}{2} - 7x\right)} = \cos{\left(\2-5x\right)}
    At this stage, instead of squaring you can write:

    \cos\left(\frac{\pi}{2}-\frac{1}{2}+7x\right) = \cos \left( 2-5x \right)

    Now the general solution for the cosine is:

     \cos{\theta} = \cos{\alpha} \Leftrightarrow \theta = 2n\pi \pm \alpha.

    From this it of course follows that

    x = \left(4n-1\right)\frac{\pi}{4}-\frac{3}{4}, ~~ \left(4n-1\right)\frac{\pi}{24}+\frac{5}{24}.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jun 2011
    From
    Bjelovar, Croatia
    Posts
    16
    Thanks
    1
    That's so much easier. Forget mine
    Last edited by Ackbeet; June 6th 2011 at 05:36 AM. Reason: Bad language.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 14th 2011, 07:07 AM
  2. Solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: August 24th 2010, 08:17 AM
  3. solving a trig equation
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: February 17th 2010, 05:29 PM
  4. Solving trig equation...
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 1st 2009, 05:43 AM
  5. need some help with solving a trig equation
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: November 16th 2008, 12:13 PM

Search Tags


/mathhelpforum @mathhelpforum