Find the slope of the tangent to the curve r=3-2sin(theta) at the point theta=r

Hello

find theta, where theta = r:

r = 3 - 2sin(r)

solutions:
theta1 = 1,16
theta2 = 3,28
theta 3 = 4,94
--------
r = 3 - 2sin(Θ)
r' = - 2cos(Θ) --> plug into equation for y': y' is the slope for any tangent at theta, and calculated by the following formula:

y' = (r' *sin(Θ) + r *cos(Θ)) / (r' *cos(Θ) - r *sin(Θ))

..... (- 2cos(Θ)*sin(Θ) + (3 - 2sin(Θ))*cos(Θ))
y' = .---------------------------------------…
......(- 2cos^2(Θ) - (3 - 2sin(Θ)*sin(Θ))


Now plug in the values for theta1, theta2, and theta 3, and find y'.
y' is then the slope of the tangents at the 3 points, where r = theta.

the slopes are:
for theta = 1,16: slope = 0,1919
for theta = 3,28: slope = 2,3298
for theta = 4,94: slope = 0,3297

Regards

We have: r = 3 - 2sin(theta)
This is an equation relating r to theta: polar co-ordinates.

Any point on this curve will have gradient dr/d(theta). The tangent at that same point will also have that gradient.

Generally then, dr/d(theta) = -2cos(theta).
When theta = r, we obtain:
dr/d(theta) = -2cos(r). This then is the slope of the tangent at the point theta = r.

There is no real solution, if I understand the problem statement correctly. The complex solution is found by solving the problem 3-2sin(theta) =d[3-2sin(theta)]/d theta = -2cos(theta)
The solutions are
theta = arccos(-(3+i)/4)
theta = arccos(-(3-i)/4)