Find the points of intersection for r = 3 + sin(theta) and r = 3 - sin(theta)!!!! PLEASE HELP!?

r = 3 + sin(t)

r = 3 - sin(t)

3 + sin(t) = 3 - sin(t)

2sin(t) = 0

sin(t) = 0

t = pi * k

r = 3 + sin(pi * k)

r = 3

(3 , pi * k) where k is an integer

r = 2 - 2cos(t)

y = r * sin(t)

x = r * cos(t)

r * sin(t) =>

2 * (1 - cos(t)) * sin(t) =>

2sin(t) - sin(2t)

y = 2sin(t) - sin(2t)

dy/dt = 2cos(t) - 2cos(2t)

r * cos(t) =>

2 * (1 - cos(t)) * cos(t) =>

2 * (cos(t) - cos(t)^2) =>

2cos(t) - 2 * (1/2) * (1 + cos(2t)) =>

2cos(t) - 1 - cos(2t)

dx/dt =>

-2sin(t) + 2sin(2t)

dy/dx =>

(dy/dt) / (dx/dt) =>

(2cos(t) - 2cos(2t)) / (2sin(2t) - 2sin(t)) =>

(cos(t) - cos(2t)) / (sin(2t) - sin(t))

t = pi/2

(cos(pi/2) - cos(pi)) / (sin(pi) - sin(pi/2)) =>

(0 - (-1)) / (0 - 1) =>

1/(-1) =>

-1

1)we can also use polar co ordinate formula to avoid lengthy calculations.

dr /d theta = 0 +2 sin theta ===> dtheta /dr = 1 / 2 sin theta = 1/ 2 * 2 sin( theta /2) cos( theta / 2)