Intersection of 2sin(x/2) and 3cos(x/3) on 0<x<4π

[stunner5000pt]

Homework Statement

Determine the points of intersection for the two given functions on the interval 0<x<4pi

$$y = 2 \sin \frac{x}{2}$$
$$y = 3 \cos \frac{x}{3}$$

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 can't seem to solve it!

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

$$\frac{e^{ix/2}} + e^{-ix/2}}{e^{ix/3}-e^{-ix/3}} = \frac{i}{2}$$

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

$$\frac{a + a^{-1}}{a^{2/3}-a^{-2/3}} = \frac{i}{2}$$

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

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!

 

[HallsofIvy]

Using exponential looks to me like the hard way to do it!
You are told that $y = 2 \sin \frac{x}{2}$ and $y = 3 \cos \frac{x}{3}$ so $2 \sin \frac{x}{2}= 3\cos\frac{x}{3}$. 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 $cos(2y)= cos^2(y)- sin^2(y)$ and $sin(2y)= 2sin(y)cos(y)$ so that $sin(3y)= sin(y+ 2y)= sin(y)cos(2y)+ cos(y)sin(2y)= sin(y)(cos^2(y)- sin^2(y))= sin(y)cos^2(y)- sin^3(y)$. The equation 2sin(3y)= 3cos(2y) becomes $2sin(y)cos^2(y)- 2cos^2(y)= 3cos^2y- 3 sin^2(y)$ which you can treat as a quadratic equation in either cos(y) or sin(y).