Results 1 to 6 of 6

Math Help - Trig equations

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    34

    Trig equations

    Hello,

    I am having a bit of trouble solving this eqn:

    tan (x) + cot (x) = 2sec (x)
    For all angles between 0 and 360 degrees.

    Thank you!

    Max
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Quote Originally Posted by darksupernova View Post
    Hello,

    I am having a bit of trouble solving this eqn:

    tan (x) + cot (x) = 2sec (x)
    For all angles between 0 and 360 degrees.

    Thank you!

    Max
    Dear darksupernova,

    tanx+cotx=2secx

    \frac{1}{sin(x)cos(x)}=2secx

    2sinx=1

    sinx=\frac{1}{2}

    Since x is inbetween 0 and 360 degrees,

    x=30^0

    Hope this will help you.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2010
    Posts
    34
    Ah very clever!

    Thank you very much!

    Im now a but stuck on this:

    tan (x) + 3cot (x) = 5sec (x)

    Very similar but the cos^2 {x) + sin^2 (X) = 1 trick doesnt work?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by darksupernova View Post
    Hello,

    I am having a bit of trouble solving this eqn:

    tan (x) + cot (x) = 2sec (x)
    For all angles between 0 and 360 degrees.

    Thank you!

    Max
    tanx+cotx=\frac{sinx}{cosx}+\frac{cosx}{sinx}=\fra  c{sinx}{sinx}\ \frac{sinx}{cosx}+\frac{cosx}{cosx}\ \frac{cosx}{sinx}

    =\frac{sin^2x+cos^2x}{sinxcosx}

    \Rightarrow\ \frac{1}{sinxcosx}=\frac{2}{cosx}\ \Rightarrow\ \frac{1}{sinx}=2\ \Rightarrow\ sinx=\frac{1}{2}

    The vertical co-ordinate of the angle is 0.5 in the unit circle, centre (0,0).
    This corresponds to two angles,

    x=sin^{-1}\left(\frac{1}{2}\right),\ x={\pi}-sin^{-1}\left(\frac{1}{2}\right)

    x=30^o,\ x=150^o
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by darksupernova View Post
    Ah very clever!

    Thank you very much!

    Im now a but stuck on this:

    tan (x) + 3cot (x) = 5sec (x)

    Very similar but the cos^2 {x) + sin^2 (X) = 1 trick doesnt work?
    \frac{sinx}{cosx}+\frac{3cosx}{sinx}=\frac{5}{cosx  }

    \frac{sin^2x+3cos^2x}{cosxsinx}=\frac{5}{cosx}

    \frac{sin^2x+cos^2x+2cos^2x}{cosxsinx}=\frac{1+2co  s^2x}{cosxsinx}=\frac{5}{cosx}

    \frac{1+2cos^2x}{sinx}=5

    2cos^2x+1=5sinx

    2-2sin^2x-5sinx+1=0

    2sin^2x+5sinx-1=0

    sinx=\frac{-5\pm\sqrt{25+8}}{4}

    x is the inverse sine of the 2 results.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Apr 2010
    Posts
    34
    Another clever little trick on line three, hadnt thought of that, thankyou very much! Now that ive seen these tricks i wont forget them!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig equations
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: March 24th 2010, 12:58 PM
  2. trig equations
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 17th 2008, 04:22 AM
  3. Trig Equations with Multiple Trig Functions cont.
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 7th 2008, 05:50 PM
  4. Trig Equations with Multiple Trig Functions help
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 6th 2008, 05:48 PM
  5. Trig Equations with Multiple Trig Functions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 6th 2008, 03:48 PM

Search Tags


/mathhelpforum @mathhelpforum