Results 1 to 4 of 4

Math Help - Trigonemetric Equation

  1. #1
    Newbie
    Joined
    Jul 2009
    Posts
    17

    Trigonemetric Equation

    cos t + 2 sec t = -3

    (t = theta)

    I need to determine all solutions of the given equation and express in radian measure.

    So far I have:

    cos^2 t + cos 0 + 2 = 0
    (cost+1)(cost+2)

    Then,
    cost=-1 cost=-2

    cos(-2) will not work, but cos(-1) = 3.14, or Pi. So the only real answer is Pi.

    Does this look correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by Snowcrash View Post
    cos t + 2 sec t = -3

    (t = theta)

    I need to determine all solutions of the given equation and express in radian measure.

    So far I have:

    cos^2 t + cos 0 + 2 = 0
    (cost+1)(cost+2)

    Then,
    cost=-1 cost=-2

    cos(-2) will not work, but cos(-1) = 3.14, or Pi. So the only real answer is Pi.

    Does this look correct?
    Not quite.

     \cos(\theta) = -1 \, \, \therefore \, \, \theta = (2k+1) \pi \, \, \forall \, k \in Z

    In other words  \theta = \text{...} -5 \pi, \, -3 \pi, \, -\pi, \, \pi, \, 3\pi, \, 5\pi \text{...} . All odd multiples of  \pi.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,139
    Thanks
    1013
    Quote Originally Posted by Snowcrash View Post
    cos t + 2 sec t = -3

    (t = theta)

    I need to determine all solutions of the given equation and express in radian measure.

    So far I have:

    cos^2 t + cos 0 + 2 = 0 don't understand this term
    (cost+1)(cost+2)

    Then,
    cost=-1 cost=-2

    cos(-2) will not work, but cos(-1) = 3.14, or Pi. So the only real answer is Pi.

    Does this look correct?
    \cos{t} + 2\sec{t} = -3

    \cos^2{t} + 2 = -3\cos{t}

    \cos^2{t} + 3\cos{t} + 2 = 0

    (\cos{t} + 1)(\cos{t}+2) = 0

    \cos{t} = -1 ... t = \pi

    \cos{t} = -2 ... no solution
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jul 2009
    Posts
    593
    Thanks
    4
    I'm not sure where you got:

    \cos^{2}(x)+\cos(x)+2=0

    It should be:

    \cos^{2}(x)+3\cos(x)+2=0

    And you have factored it correctly in any case, but if you multiply it out, you will see that you do not get your original equation.

    As for the next step:

    \cos(x)=-1
    \cos(x)=-2

    We are being asked here to evaluate WHERE cosine is -1 and where cosine is -2. We know from the domain and range of cosine (and sine) that \cos(x)=-2 is a function with no solution. However, \cos(x)=-1 has the solution of \pi. Yet, is that the ONLY solution to the problem. Remember again the graph of sine and cosine: their domain is pos/neg infinity with a repeating graph yes? This would mean there are an INFINITE many number of solutions to our equation. We then need to write our answer in a manner that makes sense. Can you figure out how to write that solution?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Which Trigonemetric Identity do I need to use?
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: August 31st 2011, 01:07 PM
  2. Replies: 1
    Last Post: April 11th 2011, 02:17 AM
  3. Trigonemetric Function - Solving
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: November 7th 2009, 07:46 PM
  4. Replies: 1
    Last Post: November 7th 2009, 12:07 AM
  5. Trigonemetric equation help
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: February 29th 2008, 09:36 AM

Search Tags


/mathhelpforum @mathhelpforum