Results 1 to 7 of 7

Math Help - 3cosx+4tanx+secx=0

  1. #1
    Junior Member jonnygill's Avatar
    Joined
    Feb 2011
    Posts
    57

    3cosx+4tanx+secx=0

    hello,

    i'm trying to work out this problem to solve for Theta...

    3cos\Theta+4tan\Theta+sec\Theta=0

    I decided it would make sense to get all trig functions to be in terms of sec.

    So after manipulating the equation i came up with...

    15sec^4\Theta-22sec^2\Theta=9

    or

    15sec^4\Theta-22sec^2\Theta-9=0

    i tried manipulating it a bit more and got...

    9cos^4\Theta+22cos^2\Theta-15=0

    i kind of keep going on circles with this one. is there something i'm missing? perhaps i approached this problem the wrong way?

    thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    use \tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta}

    and \sec(\theta)=\dfrac{1}{\cos(\theta)}

    and simplify further
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member jonnygill's Avatar
    Joined
    Feb 2011
    Posts
    57
    Quote Originally Posted by harish21 View Post
    use \tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta}

    and \sec(\theta)=\dfrac{1}{\cos(\theta)}

    and simplify further

    ok, just did that and got...

    3cos\Theta+\dfrac{4}{cos^2\Theta}+\dfrac{1}{cos\Th  eta}=0

    i'm not sure how to proceed from here
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,932
    Thanks
    334
    Awards
    1
    Quote Originally Posted by jonnygill View Post
    9cos^4\Theta+22cos^2\Theta-15=0
    Let x = cos^2(theta). Then your equation is 9x^2 + 22x - 15 = 0. Solve for x. Then use cos(theta) = sqrt(x).

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member jonnygill's Avatar
    Joined
    Feb 2011
    Posts
    57
    Quote Originally Posted by topsquark View Post
    Let x = cos^2(theta). Then your equation is 9x^2 + 22x - 15 = 0. Solve for x. Then use cos(theta) = sqrt(x).

    -Dan

    i made this substitution, and used the quadratic formula to find possible values of x. i ignored the solution where x=-6, since we will be taking the sqrt of a negative number, and i'm not interested in non-real solutions for this problem. so, i got cos\Theta=\pm\sqrt\dfrac{5}{9}.

    i solved for Theta and got... \Theta=138.1897 OR \Theta=41.8103

    i plugged both of these into the original equation and neither one worked!!

    what did i do wrong???
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by jonnygill View Post
    hello,

    i'm trying to work out this problem to solve for Theta...

    3cos\Theta+4tan\Theta+sec\Theta=0

    I decided it would make sense to get all trig functions to be in terms of sec.

    So after manipulating the equation i came up with...

    15sec^4\Theta-22sec^2\Theta=9

    or

    15sec^4\Theta-22sec^2\Theta-9=0

    i tried manipulating it a bit more and got...

    9cos^4\Theta+22cos^2\Theta-15=0

    i kind of keep going on circles with this one. is there something i'm missing? perhaps i approached this problem the wrong way?

    thanks!
    Crikey jonny!

    you really seem to like making things hard for yourself!

    Here's another way you could do it.
    Express in terms of simplest trig functions, sine and cosine..


    \displaystyle\ 3cos\Theta+4\frac{sin\Theta}{cos\Theta}+\frac{1}{c  os\Theta}=0

    obtain a common denominator, after which the numerator must be zero...

    \displaystyle\frac{3cos^2\Theta}{cos\Theta}+4\frac  {sin\Theta}{cos\Theta}+\frac{1}{cos\Theta}=0

    \displaystyle\frac{3cos^2\Theta+4sin\Theta+1}{cos\  Theta}=0

    3\left(1-sin^2\Theta\right)+4sin\Theta+1=0

    3-3sin^2\Theta}+4sin\Theta+1=0

    3sin^2\Theta-4sin\Theta-4=0

    This factors into

    \left(3sin\Theta+2)(sin\Theta-2)=0

    and one of those two factors can never be zero.
    Last edited by Archie Meade; February 24th 2011 at 02:38 PM. Reason: slight typo
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,932
    Thanks
    334
    Awards
    1
    Quote Originally Posted by jonnygill View Post
    i made this substitution, and used the quadratic formula to find possible values of x. i ignored the solution where x=-6, since we will be taking the sqrt of a negative number, and i'm not interested in non-real solutions for this problem. so, i got cos\Theta=\pm\sqrt\dfrac{5}{9}.

    i solved for Theta and got... \Theta=138.1897 OR \Theta=41.8103

    i plugged both of these into the original equation and neither one worked!!

    what did i do wrong???
    The reference angle is about 41 degrees. So far so good. Now, put that reference angle in the 4th quadrant and what do you get?

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove Cos3x=4(cos^3)x-3cosx
    Posted in the Trigonometry Forum
    Replies: 8
    Last Post: November 15th 2010, 11:57 AM
  2. Replies: 1
    Last Post: April 29th 2010, 04:28 AM
  3. 3cosx and Perpendiculars/Parallel
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 1st 2009, 10:30 AM
  4. Replies: 2
    Last Post: December 11th 2008, 03:42 PM
  5. tan²x+4tanx+1=0 help!!
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 5th 2008, 09:56 PM

Search Tags


/mathhelpforum @mathhelpforum