Results 1 to 8 of 8
Like Tree3Thanks
  • 1 Post By skeeter
  • 1 Post By princeps
  • 1 Post By Plato

Math Help - Solve giving all solutions in the interval 0 ≤ x ≤ 180

  1. #1
    Newbie
    Joined
    Mar 2012
    From
    Trinidad
    Posts
    19

    Solve giving all solutions in the interval 0 ≤ x ≤ 180

    I have to solve sec x = 4 − 2 tan^2 x, giving all solutions in the interval 0 ≤ x ≤ 180 , i have no idea where to start..... so far i changed the tan into sec^2-1, but i still cannot solve, i think they want me to use binomial expansion
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,130
    Thanks
    1011

    Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

    Quote Originally Posted by johnreal View Post
    I have to solve sec x = 4 − 2 tan^2 x, giving all solutions in the interval 0 ≤ x ≤ 180 ...
    \sec{x} = 4 - 2(\sec^2{x} - 1)

    \sec{x} = 4 - 2\sec^2{x} + 2

    2\sec^2{x} + \sec{x} - 6 = 0

    this quadratic will factor ...
    Thanks from johnreal
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2011
    From
    Crna Gora
    Posts
    420
    Thanks
    64

    Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

    Quote Originally Posted by johnreal View Post
    I have to solve sec x = 4 − 2 tan^2 x, giving all solutions in the interval 0 ≤ x ≤ 180 , i have no idea where to start..... so far i changed the tan into sec^2-1, but i still cannot solve, i think they want me to use binomial expansion
    \frac{1}{\cos x}=4-2\cdot \frac{\sin^2 x}{\cos^2 x} \Rightarrow \frac{1}{\cos x}=4-2\cdot \frac{1- \cos^2 x}{\cos^2 x} \Rightarrow

    \Rightarrow \cos x =4\cos^2 x-2+2\cos^2 x \Rightarrow 6\cos^2 x-\cos x-2=0

    Now substitute :

    \cos x=t

    and solve quadratic equation .
    Thanks from johnreal
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1

    Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

    Quote Originally Posted by johnreal View Post
    I have to solve sec x = 4 − 2 tan^2 x, giving all solutions in the interval 0 ≤ x ≤ 180 , i have no idea where to start..... so far i changed the tan into sec^2-1, but i still cannot solve, i think they want me to use binomial expansion
    Rewrite the equation: 2\sec^2(x)+\sec(x)-6=0.
    The solve 2u^2+u-6=0\text{ where }u=\sec(x)~.
    Thanks from johnreal
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2012
    From
    Trinidad
    Posts
    19

    Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

    Thanks all of you, the two different methods really helped me understand and the substitution steps cleared it all up, thank alot guys.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2012
    From
    Trinidad
    Posts
    19

    Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

    ok guys one last clarification, i now have sec x=3/2 and sec x=-2, how can i find for x.... this is the first time i'm doing these kind of questions.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2012
    From
    Trinidad
    Posts
    19

    Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

    Never mind i got it, i put sec as 1/cos then reversed it so cos=2/3 and cos=-1/2 which resulted in 48.2 and 120 degrees, thank again i think im getting the hang of this.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Mar 2012
    From
    California
    Posts
    44
    Thanks
    6

    Re: Solve giving all solutions in the interval 0 ≤ x ≤ 180

    Oops! I did this and realized you already have the solution. At any rate here it is.
    Solve giving all solutions in the interval 0  ≤ x ≤ 180-mfh-1-.jpg
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 19th 2011, 02:38 AM
  2. Replies: 3
    Last Post: July 27th 2010, 04:46 PM
  3. Solve sin(x+pi/3) = -1 for 0 ≤ x ≤2pi
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: March 28th 2010, 08:45 PM
  4. Replies: 10
    Last Post: November 15th 2009, 07:04 PM
  5. Replies: 1
    Last Post: May 27th 2008, 08:56 AM

Search Tags


/mathhelpforum @mathhelpforum