Results 1 to 6 of 6

Math Help - Two much trig :)

  1. #1
    Newbie Bobby Bob's Avatar
    Joined
    Mar 2005
    Posts
    8

    Two much trig :)

    solve 2cosX + tanX = secX where X is greater than or equal to zero and less than 2 pi. I understand that the secant is basically the inverse of the cosine, (hyp/adj), but I have no idea how to solve for this. I also don't understand how to solve an equation with 2 different trig functions in it (sin and cos).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Apr 2005
    Posts
    103
    2*cos(x) + tan(x) = sec(x)
    2*cos(x) + sin(x)/cos(x) = 1/cos(x)
    2*cos(x)^2 + sin(x) = 1
    Because sin(x)^2 + cos(x)^2 = 1,
    2*(1-sin(x)^2) + sin(x) = 1
    2 - 2*sin(x)^2 + sin(x) = 1
    1 - 2*sin(x)^2 + sin(x) = 0
    Let y = sin(x), then
    1 - 2y^2 + y = 0
    Quadratic. y = 1 or -1/2
    sin(x) = 1 or sin(x) = -1/2
    x = Pi/2 (in domain) or x = -Pi/6 (out of domain)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie Bobby Bob's Avatar
    Joined
    Mar 2005
    Posts
    8
    Thanks Paul,

    How did you use the fact that
    sin(x)^2 + cos(x)^2 = 1

    I am missing the intermediate step.

    Thank you for helping though!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Apr 2005
    Posts
    103
    Because sin(x)^2 + cos(x)^2 = 1, cos(x)^2 = 1 - sin(x)^2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    vms
    vms is offline
    Newbie vms's Avatar
    Joined
    Apr 2005
    From
    New Delhi, India
    Posts
    17
    Quote Originally Posted by paultwang
    2*cos(x) + tan(x) = sec(x)
    2*cos(x) + sin(x)/cos(x) = 1/cos(x)
    2*cos(x)^2 + sin(x) = 1
    Because sin(x)^2 + cos(x)^2 = 1,
    2*(1-sin(x)^2) + sin(x) = 1
    2 - 2*sin(x)^2 + sin(x) = 1
    1 - 2*sin(x)^2 + sin(x) = 0
    Let y = sin(x), then
    1 - 2y^2 + y = 0
    Quadratic. y = 1 or -1/2
    sin(x) = 1 or sin(x) = -1/2
    x = Pi/2 (in domain) or x = -Pi/6 (out of domain)
    Pi/2 also is not in the domain ...as tan(x) and sec(x) are not defined at x=pi/2
    hence the equation has no solutions
    method of arriving at the solution is correct but.... small error in concluding the answer
    Last edited by vms; April 14th 2005 at 11:01 PM. Reason: small correction
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    -pi/6 was found to be one of the two possible values of angle x. And it was said to be not in the domain of x.

    The domain of x was mentioned to be
    x >= 0, and x < 2pi
    Rewriting that,
    0 <= x < 2pi
    which means x can be from 0 to almost 2pi. Meaning x can be 0 but x cannot be 2pi.

    sin x = -1/2
    x = arcsin(-1/2) = -pi/6
    that is if we measure x clockwise. But even here -pi/6 is in the 4th quadrant.

    What if we measure by the usual way---by counterclockwise? Which is intended by the given domain anyway?
    What is x then when sin x = -1/2?

    Sine is negative in the 3rd and 4th quadrants.
    If sin x = 1/2, then x = pi/6.
    How would you place this pi/6 then in the 3rd and 4th quadrants so that you can get sin x = -1/2?

    You add pi/6 to pi, and you subtract pi/6 from 2pi.
    Or,
    x = pi +pi/6 = 7pi/6 ....answer.
    x = 2pi -pi/6 = 11pi/6 ....answer.

    In degrees,
    pi/6 = 30 degrees; pi = 180 deg; and 2pi = 360 deg
    So,
    x = 180 +30 = 210 deg ....answer.
    x = 360 -30 = 330 deg ....answer.

    Those two values of x are in the given domain.
    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, 08:00 PM
  2. Replies: 7
    Last Post: April 15th 2010, 09:12 PM
  3. Replies: 6
    Last Post: November 20th 2009, 05:27 PM
  4. Replies: 1
    Last Post: July 24th 2009, 03:29 AM
  5. Trig Equations with Multiple Trig Functions cont.
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 7th 2008, 06:50 PM

Search Tags


/mathhelpforum @mathhelpforum