How to find intersection points of 2 curves?

Question

I'm given r = 2 (circle with radius 2)
r^2 = 9sin(2theta) [lemniscate with "radius" 3]

I need to find 4 points of intersection, although I'm not sure how to without calculus.

Perhaps I'm being unclear...
I got to the point 4/9 = sin(2theta), but I don't know where to go from here.

Answer 1

r = 2
r^2 = 9sin(2t)
2^2 = 9sin(2t)
4 = 9sin(2t)
sin(2t) = 4/9

So that that extent, you are correct. To continue, you need trigonometry rather than calculus. Normally, we limit theta (which I've rewritten as t for convenience) to the range 0 <= t < 2*pi. The arcsine function, on the other hand, returns values -pi/2 <= t <= pi/2. For a positive value y, t = arcsine(y) will be positive. However, 2t = arcsine(4/9) ==> t = (1/2)arcsine(4/9) isn't the only value of t for which sin(2t) = 4/9. This is because sin(pi - arcsine(4/9)) = 4/9 as well; this angle corresponds to the angle arcsine(4/9) as a reference angle in the second quadrant, where sine is also positive. Therefore, t = (1/2)(pi - arcsine(4/9)) is also a correct answer.

But, wait, there's more! If you double these angles, you do get an angle whose sine is 4/9. Now, what what if you double an angle that is pi greater than each of these? Well, you add 2*pi to the angle. But 2*pi is a full circle, meaning you get the exact same angle from a geometric perspective. So the full set of values for t is as follows:

t = (1/2)arcsine(4/9)
t = (1/2)arcsine(4/9) + pi
t = (1/2)(pi - arcsine(4/9))
t = (1/2)(pi - arcsine(4/9)) + pi

And for all of these values of t, the corresponding value for r is r = 2 because it was given as constant.

Answer 2

set both equations equal to each other and then solve for r