Results 1 to 5 of 5

Math Help - Need help with a Trig equation

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    11

    Need help with a Trig equation

    2sin\theta  cos\theta-cos\theta=0

    This is what I did

    First I took cos\theta to other side resulting in 2sin\theta  cos\theta=cos\theta
    Then I divided by cos\theta resulting in 2sin\theta =1 then I divided by 2 sin\theta = \frac{1}{2}
    I am sure it's incorrect.. Can someone please help me
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

    Dividing by \cos\theta makes you lose solutions.

    You must factor.

    2\sin\theta  \cos\theta-\cos\theta=0

    \cos\theta (2\sin\theta-1)=0

    leading to \cos\theta = 0 or \sin\theta = \frac{1}{2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by daunder View Post
    2sin\theta  cos\theta-cos\theta=0

    This is what I did

    First I took cos\theta to other side resulting in 2sin\theta  cos\theta=cos\theta
    Then I divided by cos\theta resulting in 2sin\theta =1 then I divided by 2 sin\theta = \frac{1}{2}
    I am sure it's incorrect.. Can someone please help me
    That would give you some solutions but not all of them. You can get them all by factorising:

    cos(\theta)(2sin(\theta)-1) = 0

    Then you can solve cos(\theta) = 0 to give \theta = \frac{\pi}{2} \text { , } \frac{3\pi}{2}

    Then solve 2sin(\theta)-1 = 0

    \theta = \frac{\pi}{6} \text { , } \frac{5\pi}{6}

    note: I have worked out the solutions of 0 \leq \theta \leq  2\pi
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2009
    Posts
    11
    Thanks for the quick reply
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by daunder View Post
    2sin\theta  cos\theta-cos\theta=0

    This is what I did

    First I took cos\theta to other side resulting in 2sin\theta  cos\theta=cos\theta
    Then I divided by cos\theta resulting in 2sin\theta =1 then I divided by 2 sin\theta = \frac{1}{2}
    I am sure it's incorrect.. Can someone please help me
    You can't divide by \cos(\theta) as it can be equal to 0.

    Here is how to do.

    factorise by cos(theta) :

    \cos \theta (2 \sin \theta-1)=0

    the product ab=0 if and only if a=0 or b=0

    ----> \cos \theta=0 or \sin \theta=\frac 12


    For the first one... there are infinitely many solutions !
    we know that \cos \frac \pi 2=0
    but since cos(-x)=cos(x), we know that \cos \left(-\frac \pi 2\right)=0
    moreover, cos is a 2pi-periodic function.
    Hence \boxed{\cos \left(\frac \pi 2 +2k\pi\right)=\cos \left(-\frac \pi 2 +2k\pi\right)=0}
    these give all the possible solutions for cos(theta)=0

    (in general, if a is such that cos(a)=b, then the solutions to cos(x)=b are in the form a+2kpi or -a+2kpi, for any integer k)


    For the second one, there are infinitely many solutions too.
    we know that \sin\left(\frac \pi 3\right)=0
    but we know that \sin(\pi-x)=\sin(x). Thus \sin\left(-\frac \pi 3+\pi\right)=0
    again, sin is a 2pi-periodic function.
    Hence \sin\left(2k\pi\right)=\sin\left(-\frac \pi 3+\pi+2k\pi\right)=0
    this can be rewritten :
    \boxed{\sin\left(2k\pi\right)=\sin\left(-\frac \pi 3+(2k+1)\pi\right)=0}


    Finally, the solutions are :
    S=\left\{\tfrac{\pi}{2}+2k \pi \big| k \in \mathbb{Z}\right\} \cup \left\{-\tfrac{\pi}{2}+2k \pi \big| k \in \mathbb{Z}\right\} \cup \left\{2k \pi \big| k \in \mathbb{Z}\right\} \cup \left\{-\tfrac \pi 3+(2k+1)\pi \big| k \in \mathbb{Z}\right\}

    does it look good to you ?




    Edit : woops, a bit late
    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