Results 1 to 4 of 4

Math Help - Trig equations: cos4x=sin2x

  1. #1
    RAz
    RAz is offline
    Junior Member RAz's Avatar
    Joined
    May 2008
    From
    Canada
    Posts
    54

    Trig equations: cos4x=sin2x

    I have to solve cos4x=sin2x where x is an element of 0, 360.

    The problem, for me, is cos4x. I haven't proved this myself, but a quick search gave me that cos4x= 8cos^4x-8cos^2x+1

    By going along the lines of 8cos^4x-8cos^2x+1=2sinxcosx (see original question) I managed to arrive at sinx= \frac{1}{3}

    This does not give an exact value and I was wondering if I had done this correctly. Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008
    Quote Originally Posted by RAz View Post
    I have to solve cos4x=sin2x where x is an element of 0, 360.

    The problem, for me, is cos4x. I haven't proved this myself, but a quick search gave me that cos4x= 8cos^4x-8cos^2x+1

    By going along the lines of 8cos^4x-8cos^2x+1=2sinxcosx (see original question) I managed to arrive at sinx= \frac{1}{3}

    This does not give an exact value and I was wondering if I had done this correctly. Thanks!
    \cos{4x} = \sin{2x}

    \cos^2{2x} - \sin^2{2x} = \sin{2x}

    1 - \sin^2{2x} - \sin^2{2x} = \sin{2x}

    1 - 2\sin^2{2x} = \sin{2x}

    0 = 2\sin^2{2x} + \sin{2x} - 1.


    You now have a quadratic equation in \sin{2x}.

    So let \sin{2x} = X, so that you can see the Quadratic:


    0 = 2X^2 + X - 1

    0 = 2X^2 + 2X - X - 1

    0 = 2X(X + 1) - 1(X + 1)

    0 = (X + 1)(2X - 1).


    So X = -1 or X = \frac{1}{2}.


    This means \sin{2x} = -1 or \sin{2x} = \frac{1}{2}.


    Case 1:

    \sin{2x} = -1

    2x = 270^{\circ} + 360^{\circ}n, where n is an integer.

    x = 135^{\circ} + 180^{\circ}n, where n is an integer.


    Case 2:

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

    Since sine is positive in the first and second quadrants:

    2x = \left\{30^{\circ}, 180^{\circ} - 30^{\circ} \right\} + 360^{\circ} n

    2x = \left\{ 30^{\circ}, 150^{\circ}\right\} + 360^{\circ} n

    x = \left \{ 15^{\circ}, 75^{\circ} \right\} + 180^{\circ} n.



    So putting it together:

    x = \left\{ 15^{\circ}, 75^{\circ}, 135^{\circ}\right\} + 180^{\circ}n.


    Upon substituting n = 0 and n = 1, we find all the possible solutions in the domain 0^{\circ} \leq x \leq 360^{\circ}:


    x = \left\{ 15^{\circ}, 75^{\circ}, 135^{\circ}, 195^{\circ}, 255^{\circ}, 315^{\circ} \right\}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    RAz
    RAz is offline
    Junior Member RAz's Avatar
    Joined
    May 2008
    From
    Canada
    Posts
    54
    Thanks a lot, I appreciate it. Very clear.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Just to give an alternative way

    \cos{4x} = \sin{2x}

    \cos{4x} = \cos{(90^{\circ}-2x)}

    Case 1:
    4x = 90^{\circ}-2x+ 360^{\circ} n

    6x = 90^{\circ} + 360^{\circ} n

    x = 15^{\circ} + 60^{\circ} n

    x = \left\{ 15^{\circ}, 75^{\circ}, 135^{\circ}, 195^{\circ},  255^{\circ}, 315^{\circ} \right\}


    Case 2:
    4x = -(90^{\circ}-2x) + 360^{\circ} n

    4x = -90^{\circ}+2x + 360^{\circ} n

    2x = -90^{\circ} + 360^{\circ} n

    x = -45^{\circ} + 180^{\circ} n

    x = \left\{135^{\circ}, 315^{\circ} \right\}


    Therefore:
    x = \left\{ 15^{\circ}, 75^{\circ}, 135^{\circ}, 195^{\circ},   255^{\circ}, 315^{\circ} \right\}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove : cosx,cos2x,cos4x,cos6x
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: August 5th 2009, 09:30 AM
  2. solving trig functions sin2x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 22nd 2009, 12:31 AM
  3. Trig Proof/Verification : cot 2x = (1+cos4x)/sin4x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 10th 2008, 08:25 PM
  4. Replies: 4
    Last Post: May 2nd 2008, 12:12 PM
  5. cos4x
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: March 11th 2008, 11:35 PM

Search Tags


/mathhelpforum @mathhelpforum