Find the coordinates [r, theta] of all the points of intersection: r=2sin(theta) and r=1?

Answer 1

Set the two expressions for r equal to each other:
1 = 2sin(θ)
1/2 = sin(θ)

So θ = 30, 150. Or in radians: π/6, 5π/6. This makes the intersection points (1, π/6) and (1, 5π/6).

We could generalize the solution of 1/2 = sin(θ) to include angles like -7π/6 or 13π/6, but that would be needlessly redundant because they'd just be one of the same two coordinate points listed above.

Answer 2

Brats almost has it right.

There are only two points because r = 2sinθ is a circle centre (0,1) radius 1 and r = 1 is a circle centre (0,0) and radius 1.

These only meet at (1, π/6) and (1, 5π/6).

His solution is correct for sinθ = ½ but adding 2π will not produce any more points.

Answer 3

since r = 1, we have 2 sin theta is 1 or sin theta is 1/2

so theta is n pi + (-1)^n * pi / 6

hence the points are (1, pi/6) and (1, 5 pi/6)
other points can be achieved by adding 2n pi to theta

Answer 4

2sint=1 so sin t =1/2 and t= pi/6+2kpi and t = 5pi/6+2kpi
so all the pointa are
(1,pi/6+2kpi) and (1,5pi/6+2kpi)