Solving a trigonometric equation

[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).

 

[Architeuthis Dux]

I would say that your approach is definitely valid and shows good problem-solving skills. However, there may be a simpler approach to solving this trigonometric equation. One possible way is to use the double angle formula for sine and the half angle formula for cosine to rewrite the given functions as:

y = 2sin(x/2) = 2(2sin^2(x/4)-1) = 4sin^2(x/4)-2

y = 3cos(x/3) = 3(2cos^2(x/6)-1) = 6cos^2(x/6)-3

Now, we can equate these two functions and solve for x:

4sin^2(x/4)-2 = 6cos^2(x/6)-3

4sin^2(x/4)-6cos^2(x/6) = -1

Using the double angle formula for sine and the half angle formula for cosine, we can simplify this equation to:

4(1-cos(x/2))-6(1+cos(x/3)) = -1

4-4cos(x/2)-6-6cos(x/3) = -1

-4cos(x/2)-6cos(x/3) = 11

Now, we can use the product-to-sum formula for cosine and simplify the equation further:

-4(cos(x/2+cos(x/3)) = 11

cos(x/2+cos(x/3)) = -11/4

We know that the cosine function has a maximum value of 1 and a minimum value of -1, so there are no real solutions for this equation. This means that there are no points of intersection for the two given functions on the interval 0<x<4pi. It is possible that there may be complex solutions, but that is beyond the scope of this problem.

In conclusion, it is always important to try different approaches when solving equations, and your attempt shows good problem-solving skills. However, in this case, there may be a simpler approach using trigonometric identities. It is always a good idea to review and familiarize yourself with these identities to make problem-solving easier.