Results 1 to 6 of 6

Math Help - Solving Trig Equations Help

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    2

    Solving Trig Equations Help

    I'm having trouble with these problems

    Question is: "Solve the following equations for 0<X<360

    1) Sin(2X)=Cos(X)

    2)1+Sin(x)=Cos^2(x)

    If anyone could help, I'd be really grateful.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by LexiLou View Post
    I'm having trouble with these problems

    Question is: "Solve the following equations for 0<X<360

    1) Sin(2X)=Cos(X)

    2)1+Sin(x)=Cos^2(x)

    If anyone could help, I'd be really grateful.
    Hi LexiLou,

    \sin 2x=2\sin x \cos x, so

    [1] \sin 2x=\cos x

    2\sin x \cos x=\cos x

    2\sin x = 1

    \sin x =\frac{1}{2}

    x=\{30^{\circ}\:,\:150^{\circ}\}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    [2] 1+ \sin x= \cos^2x

    Identity: cos^2 x = 1- \sin^2 x

    1+\sin x=1-\sin^2 x

    \sin^2 x+ \sin x=0

    \sin x(\sin x + 1)=0

    \sin x=0 \ \ or \ \ \sin x = -1


    x=\{0^{\circ}\:,\: 180^{\circ}\:,\:270^{\circ}\}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by masters View Post
    Hi LexiLou,

    \sin 2x=2\sin x \cos x, so

    [1] \sin 2x=\cos x

    2\sin x \cos x=\cos x

    2\sin x = 1

    \sin x =\frac{1}{2}

    x=\{30^{\circ}\:,\:150^{\circ}\}
    It is better to factorise these and when you divide through by \cos(x) you potentially divide by 0.


    2\sin(x) \cos(x)-\cos(x)=0

    \cos(x)[2\sin(x)-1]= 0

    Therefore in addition \cos(x)=0 also applies. This means that  x = \{90^{\circ}\:,\: 270^{\circ}\}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by e^(i*pi) View Post
    It is better to factorise these and when you divide through by \cos(x) you potentially divide by 0.


    2\sin(x) \cos(x)-\cos(x)=0

    \cos(x)[2\sin(x)-1]= 0

    Therefore in addition \cos(x)=0 also applies. This means that  x = \{90^{\circ}\:,\: 270^{\circ}\}
    Thanks, e^(i*pi). Got in a hurry! Thanks for cleaning up after me.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2010
    Posts
    2
    Thanks so much, that helped a lot.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving Trig Equations
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: March 20th 2011, 07:46 PM
  2. Solving Trig Equations
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: September 5th 2009, 07:34 PM
  3. Solving Trig Equations
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: February 25th 2009, 06:42 PM
  4. Solving trig Equations
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 5th 2008, 09:15 PM
  5. solving trig equations
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 14th 2008, 06:03 PM

Search Tags


/mathhelpforum @mathhelpforum