Results 1 to 5 of 5

Math Help - Trig problem

  1. #1
    Newbie
    Joined
    Mar 2007
    Posts
    3

    Trig problem

    Hi - I have been worrking on this problem for hours - and just am at a road block. Can anyone help me????? It would be so appreciated. This is my last problem and I just can't seem to get it started....

    Find all soluctions of the equations: (a) SinX=-1 (b) Cos u + Sin 2u = 0 (c) 2Sint Cost -Sint = 0 (d) (0, 360)

    any help with these would be greatly appreciated....

    thanks
    silly_girl
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,933
    Thanks
    336
    Awards
    1
    Quote Originally Posted by silly_girl View Post
    Find all soluctions of the equations:
    (a) SinX=-1
    As I noted in another thread, Sin(x) and sin(x) are technically two different functions. The common trig functions are written without capital letters.

    sin(x) = -1

    We know that, for 0 <= x < 2(pi), that sin(x) = -1 for only x = 3(pi)/2. This solution is periodic with a period of 2(pi), so the solution will be:
    x = 3(pi)/2 + 2(pi)*n, where n is any integer.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,933
    Thanks
    336
    Awards
    1
    Quote Originally Posted by silly_girl View Post
    (b) Cos u + Sin 2u = 0
    Note that sin(2u) = 2*sin(u)*cos(u) so:
    cos(u) + 2*sin(u)*cos(u) = 0 <-- Factor the common cos(u):

    cos(u)*[1 + 2*sin(u)] = 0

    Now either
    cos(u) = 0 ==> u = (pi)/2, 3(pi)/2

    or

    1 + 2*sin(u) = 0 ==> sin(u) = -1/2 ==> u = 7(pi)/6, 11(pi)/6

    Again, each of these solutions are periodic, so the final solution is:
    u = (pi)/2 + 2(pi)*n, 7(pi)/6 + 2(pi)*n, 3(pi)/2 + 2(pi)*n, 11(pi)/6 + 2(pi)*n, where n is any integer.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,933
    Thanks
    336
    Awards
    1
    Quote Originally Posted by silly_girl View Post
    (c) 2Sint Cost -Sint = 0
    Similar to b):
    sin(t)*[2*cos(t) - 1] = 0

    sin(t) = 0 ==> t = 0, (pi)

    2*cos(t) - 1 = 0 ==> cos(t) = 1/2 ==> t = (pi)/3, 5(pi)/3

    So for 0 <= t < 2(pi)

    t = 0, (pi)/3, (pi), 5(pi)/3
    (You can add the 2(pi)*n's this time.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,933
    Thanks
    336
    Awards
    1
    Quote Originally Posted by silly_girl View Post
    (d) (0, 360)
    What's this question again?

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with Trig problem
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 16th 2011, 02:25 PM
  2. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 14th 2011, 07:07 AM
  3. Replies: 3
    Last Post: January 2nd 2011, 08:20 PM
  4. Trig problem
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 14th 2009, 11:34 AM
  5. Trig Problem
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 21st 2008, 07:45 PM

Search Tags


/mathhelpforum @mathhelpforum