1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving a trigonometric equation

  1. Jan 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine the points of intersection for the two given functions on the interval 0<x<4pi

    [tex] y = 2 \sin \frac{x}{2} [/tex]
    [tex] y = 3 \cos \frac{x}{3} [/tex]

    2. The attempt at a solution

    Well i tried graphing it and found out that the solution must lie somewhere between pi and 2pi and that there is only one solution on this interval.
    But i cant seem to solve it!

    i tried using the complex exponential form of sine and cosine and got

    [tex] \frac{e^{ix/2}} + e^{-ix/2}}{e^{ix/3}-e^{-ix/3}} = \frac{i}{2} [/tex]

    I tried substituting e^ix/2 = a and got this

    [tex] \frac{a + a^{-1}}{a^{2/3}-a^{-2/3}} = \frac{i}{2} [/tex]

    simplifying a bit
    [tex] a^{8/3} + a^{2/3} = \frac{i}{2} (a^{7/3} - a^ {3/3}} [/tex]

    and then make another substitution...
    actally before i keep going i must ask if my way is unnecessarily longwinded...

    is there something 'obvious' in the two functions where i can use some trig identities to simplify?

    Thanks for your help!
  2. jcsd
  3. Jan 1, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Using exponential looks to me like the hard way to do it!
    You are told that [itex] y = 2 \sin \frac{x}{2} [/itex] and [itex] y = 3 \cos \frac{x}{3} [/itex] so [itex]2 \sin \frac{x}{2}= 3\cos\frac{x}{3}[/itex]. Since I personally feel that "multiples" of angles are easier to work with than "fractions", I might let y= x/6 so the equation becomes 2 sin(3y)= 3 cos(2y). Now, I recall that [itex]cos(2y)= cos^2(y)- sin^2(y)[/itex] and [itex]sin(2y)= 2sin(y)cos(y)[/itex] so that [itex]sin(3y)= sin(y+ 2y)= sin(y)cos(2y)+ cos(y)sin(2y)= sin(y)(cos^2(y)- sin^2(y))= sin(y)cos^2(y)- sin^3(y)[/itex]. The equation 2sin(3y)= 3cos(2y) becomes [itex]2sin(y)cos^2(y)- 2cos^2(y)= 3cos^2y- 3 sin^2(y)[/itex] which you can treat as a quadratic equation in either cos(y) or sin(y).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Solving a trigonometric equation