Results 1 to 12 of 12

Math Help - Help with Trig Equation!

  1. #1
    Member
    Joined
    Apr 2008
    Posts
    107

    Trig Equation! Help

    Does anyone know how to do this? I have no idea how to even start this. Thanks a lot

    2csc^2 x = 3cot^2 x - 1
    sin^2x + sin x - 1 = 0
    sin x + cos x = 0
    Last edited by NeedHelp18; May 7th 2008 at 01:13 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,846
    Thanks
    321
    Awards
    1
    Quote Originally Posted by NeedHelp18 View Post
    Does anyone know how to do this? I have no idea how to even start this. Thanks a lot

    2csc^2 x = 3cot^2 x - 1
    2~csc^2(x) = 3~cot^2(x) - 1[/tex]

    \frac{2}{sin^2(x)} = \frac{3~cos^2(x)}{sin^2(x)} - 1

    2 = 3~cos^2(x) - sin^2(x)

    and since cos^2(x) = 1 - sin^2(x)

    2 = 3 - 3~sin^2(x) - sin^2(x)

    -1 = -2~sin^2(x)

    You can finish from here.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by NeedHelp18 View Post
    Does anyone know how to do this? I have no idea how to even start this. Thanks a lot

    2csc^2 x = 3cot^2 x - 1
    sin^2x + sin x - 1 = 0
    sin x + cos x = 0
    For the first problem, write it as \frac{2}{\sin ^2x} = \frac{3 \cos ^2x}{\sin ^2x} - \frac{\sin ^2x}{\sin ^2x}. Then, as long as sin x isn't 0, you have 2 = 3 \cos ^2x - \sin ^2x, which you can then arrange as:

    2 = 3 \cos ^2x - (1 - \cos ^2x)

    2 = 3 \cos ^2x - 1 + \cos ^2x

    3 = 4 \cos ^2x

    \cos ^2x = \frac{3}{4}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,846
    Thanks
    321
    Awards
    1
    Quote Originally Posted by NeedHelp18 View Post
    sin^2x + sin x - 1 = 0
    Let y = sin(x). This transforms the equation to
    y^2 + y - 1 = 0

    Solve for y and then set that equal to sin(x).

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,846
    Thanks
    321
    Awards
    1
    Quote Originally Posted by NeedHelp18 View Post
    sin x + cos x = 0
    If you honestly can't get this one you are thinking too hard about it.
    sin(x) + cos(x) = 0

    sin(x) = -cos(x)

    \frac{sin(x)}{cos(x)} = -1

    tan(x) = -1
    and go from there.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by NeedHelp18 View Post
    Does anyone know how to do this? I have no idea how to even start this. Thanks a lot

    2csc^2 x = 3cot^2 x - 1
    sin^2x + sin x - 1 = 0
    sin x + cos x = 0
    For the second problem, substitute \sin x = z and then solve the equation z^2 + z - 1 = 0 using the quadratic formula. Once you get the two values for z, see if either of them make sense for \sin x.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Apr 2008
    Posts
    107
    Thanks! I was definitely trying too hard lol
    My teacher refused to show our class how to do them because she got mad at us for talking soo much, even though i wasnt talking at all lol
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Apr 2008
    Posts
    107
    okay so i tried to do this one and i got half right.
    i got pi/3 and 5pi/3.
    why does the book also say that i was supposed to get 0 and pi??

    here it is:

    2cos^2 x - cos x = 2 - sec x
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,846
    Thanks
    321
    Awards
    1
    Quote Originally Posted by NeedHelp18 View Post
    okay so i tried to do this one and i got half right.
    i got pi/3 and 5pi/3.
    why does the book also say that i was supposed to get 0 and pi??

    here it is:

    2cos^2 x - cos x = 2 - sec x
    New questions go in new threads.

    2~cos^2(x) - cos(x) = 2 - \frac{1}{cos(x)}

    2~cos^3(x) - cos^2(x) - 2~cos(x) + 1 = 0

    Again, letting y = cos(x) and solving the resulting polynomial I get that
    cos(x) = 1, cos(x) = -1, cos(x) = 1/2

    -Dan
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Apr 2008
    Posts
    107
    Quote Originally Posted by topsquark View Post
    New questions go in new threads.

    2~cos^2(x) - cos(x) = 2 - \frac{1}{cos(x)}

    2~cos^3(x) - cos^2(x) - 2~cos(x) + 1 = 0

    Again, letting y = cos(x) and solving the resulting polynomial I get that
    cos(x) = 1, cos(x) = -1, cos(x) = 1/2

    -Dan
    how do you do this part?
    im trying it but nothing.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Jan 2008
    Posts
    588
    Thanks
    87
    He means substituting y = cos(x)

      2 cos^3(x)-cos^2(x)-2cos(x)+1 = 0 \Rightarrow<br />
2y^3-y^2-2y+1 = 0

    and then solving that equation.

    To solve a polynomial, find a solution to that equation (in this case, y = 1 satisfies the equation) Therefore its a solution to the polynomial and (y - 1) is a factor. Divide the polynomial by that factor, and you should get a quadratic. Apply the quadratic formula to that to get the other two answers.

    Remember, when you have your solutions, don't forget to change your y back into cos(x)

    For a more indepth look at solving polynomails look at this site Solving Polynomial Equations
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Apr 2008
    Posts
    65
    so it becomes
    2y^3 - y^2 - 2y + 1 = 0

    solve for y as you would a polynomial then sub back into the y

    i get

    2 cos theta -1 = 0

    and

    cos theta + 1 = 0

    and

    cos theta -1 = 0

    depends what range the solution is required

    between 0 and 2 pi (inclusive)

    i get: 0, pi/3, pi, 5pi/3, 2pi
    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. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 14th 2011, 07:07 AM
  3. Trig Equation with varied trig functions
    Posted in the Trigonometry Forum
    Replies: 12
    Last Post: April 12th 2010, 10:31 AM
  4. Replies: 1
    Last Post: July 24th 2009, 03:56 AM
  5. Replies: 1
    Last Post: July 24th 2009, 02:29 AM

Search Tags


/mathhelpforum @mathhelpforum