How can I find all the solutions of this?-VBForums
Results 1 to 2 of 2

Thread: How can I find all the solutions of this?

  1. #1

    Thread Starter
    New Member
    Join Date
    Feb 2012

    Post How can I find all the solutions of this?

    Okay, so I'm doing this question and have no idea how to find all the solution for cosΘ = 1/√2 when the domain is 0≤Θ≤2π (that's a pi symbol)
    Here's what I have worked out so far.
    cosΘ = 1/√2
    = π/4
    cos is positive in the 1st and 4th quadrants.

    I know the second answer is 7π/4. How did the 7 get in the result? Is there an equation I have to do to get it?

    Also, I am having trouble with cosΘ = 0 when the domain is the same as above.

    Please help!
    Thank you.

  2. #2
    Only Slightly Obsessive jemidiah's Avatar
    Join Date
    Apr 2002

    Re: How can I find all the solutions of this?

    Imagine the graph of the cosine function between 0 and 2*pi. Draw horizontal lines through the graph. The graph is symmetric about x=pi, so wherever the horizontal line intersects the graph, it must intersect the graph on the other side of x=pi as well. As an example, take the horizontal line y=1. In the domain 0 <= x <= pi, cos(x) = 1 only for x=0, so in the domain pi <= x <= 2pi, cos(x) = 1 only for x = pi + (pi - 0) = 2pi.

    In your situation you should know (since it's one of the special angles) that cos(pi/4) = 1/sqrt(2) = sqrt(2)/2. Since 0 <= pi/4 <= pi, this solution can be reflected through x=pi to give a second solution, pi + (pi - pi/4) = 4pi/4 + 3pi/4 = 7pi/4.

    More generally, suppose -1 < c <= 1. From the shape of the graph (in particular, the fact that it decreases on [0, pi] from 1 to -1 and mirrors itself after that), we have the following: there is a unique point where the horizontal line y=c intersects the graph cos(x) in the domain 0 <= x <= pi, i.e. there is a unique solution (call it t) to the equation cos(x) = c when we restrict x to the range 0 <= x <= pi. Moreover, there is a corresponding unique solution t' to this equation in the range pi <= x <= 2pi. The two solutions are related by the fact that the distance from pi to t is the same as the distance from t' to pi, since the graph mirrors about x=pi. Algebraically, pi - t = t' - pi.

    Given a solution t, we can compute t' since t' = pi + (pi - t). For c = 1/sqrt(2), the above holds, and we know that t=pi/4, so t'=pi + (pi - pi/4) = 7pi/4 as above. Moreover these are the only solutions on [0, 2pi], since t is unique on [0, pi] and t' is unique on [pi, 2pi].
    The time you enjoy wasting is not wasted time.
    Bertrand Russell

    <- Remember to rate posts you find helpful.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts


Click Here to Expand Forum to Full Width

Survey posted by VBForums.