Results 1 to 5 of 5

Math Help - Trig Help.

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    42

    Trig Help.

    I need help with the following equations to solve between 0 and 360:

    1. \sin2x = \frac{1}{2}

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

    3.  \cos(x-30) = 1

    4. \cos^2x = \frac{3}{2}

    For 1, I've obtained x = 45, 225, 315, 360.
    For 2, I've 30 and 75.
    For 3, hopeless.
    For 4, no idea.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2009
    Posts
    49
    sin(2x)=1/2
    let b=2x
    values of b that give +1/2 are 30 and 150
    so sinb=1/2 where b =30,150
    b=2x implies 2x=30 and 2x = 150

    sin^2 x =1/2 is the same as (sinx)^1/2 =1/2,
    thus sinx = sqrt(1/2)
    values of x that give +1/sqrt(2) are 45 and 135

    cos(x-30)=1
    let x-30=b
    so cosb=1
    values of b that give +1 are 0 and 360
    this means x-30=0 and x-30=360

    cos^2 x =3/2
    this is the same as cosx=sqrt(3/2)
    but sqrt(3/2) is > than 1.
    cosx has a range of 1>y>-1
    so there is no solution
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    42
    Quote Originally Posted by purebladeknight View Post
    cos^2 x =3/2
    this is the same as cosx=sqrt(3/2)
    but sqrt(3/2) is > than 1.
    cosx has a range of 1>y>-1
    so there is no solution
    Hi, thanks. What about if it was \cos^2x = \frac{3}{4}? Would it have solutions?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2009
    Posts
    42
    Got it now! Can I thank myself? lol.

    Thanks for the help Purebladeknight.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello Captcha
    Quote Originally Posted by Captcha View Post
    I need help with the following equations to solve between 0 and 360:

    1. \sin2x = \frac{1}{2}

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

    ...
    I have just a couple of things to add to questions 1 and 2.

    1. If 0^o\le x \le 360^o, then 0^o\le 2x \le 720^o. So you will need to say
    2x = 30^o, 150^o, 390^o, 510^o

    \Rightarrow x = 15^o, 75^o, 195^o, 255^o
    2. Don't forget the \pm sign when you take the square root. So
    \sin^2 x = \frac12

    \Rightarrow \sin x = \pm \frac{1}{\sqrt2}

    \Rightarrow x = 45^o, 135^o, 225^o, 315^o
    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compute Trig Function Values, Solve Trig Equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 8th 2011, 07:00 PM
  2. Replies: 7
    Last Post: April 15th 2010, 08:12 PM
  3. Replies: 6
    Last Post: November 20th 2009, 04:27 PM
  4. Replies: 1
    Last Post: July 24th 2009, 02:29 AM
  5. Replies: 2
    Last Post: April 21st 2006, 03:04 PM

Search Tags


/mathhelpforum @mathhelpforum