Calculus of a Single Variable, Chapter 10, 10.4, Section 10.4, Problem 67

r=2csc theta+3
To solve, express the polar equation in parametric form. To convert it to parametric equation, apply the formula

x = rcos theta
y=r sin theta

Plugging in r=2csctheta +3 , the formula becomes:
x = (2csctheta +3)cos theta
x= 2cot theta + 3cos theta
y =(2csc theta+3)sin theta
y=2+3sintheta
So the equivalent parametric equation of r=2csctheta +3 is:
x=2cot theta +3cos theta
y=2+3sin theta
Then, take the derivative of x and y with respect to theta.
dx/(d theta ) = 2*(-csc^2 theta) + 3*(-sin theta)
dx/(d theta)=-2csc^2 theta -3sin theta
dy/(d theta)=3costheta
Take note that the slope of the tangent is equal to dy/dx.
m= (dy)/(dx)
To get the dy/dx of a parametric equation, apply the formula:
dy/dx = (dy/(d theta))/(dx/(d theta))
When the tangent line is horizontal, the slope of the tangent is zero.
0 = (dy/(d theta)) / (dx/(d theta))
This implies that the polar curve will have a horizontal tangent when numerator is zero. So set the derivative of y equal to zero.
dy/(d theta) = 0
3cos theta =0
cos theta=0
theta =pi/2, (3pi)/2
So the polar curve have a horizontal tangents at:
theta_1 = pi/2 + 2pin
theta_2 = (3pi)/2 + 2pi n
where n is any integer.
To determine the points (r, theta) , plug-in the values of theta to the polar equation.
r=2csc theta + 3
theta=pi/2+2pin ,
r=2csc(pi/2+2pin)+3= 2csc(pi/2)+3=2*1+3=5
theta=(3pi)/2+2pin ,
r=2csc((3pi)/2+2pin)+3= 2csc((3pi)/2)+3=2*(-1)+3=1
Therefore, the polar curve has horizontal tangents at points (5, pi/2+2pin) and (1, (3pi)/2+2pin) .