`r=2csctheta+3` Find the points of horizontal tangency (if any) to the polar curve.

`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...

`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)` .