Quick and easy-to-answer question about polar coordinates ( EZ 10 )

Okay, I have r ^ 2 = 36 cos (Q)

( Q suffice as theta)

I know it has symmetry about both axises and about the origin.

My Question:::> How would I solve this equation so that I can graph it OR how would I make a table to graph individual points [ similar to the way the rectangular coord. sys works x l f(x) ]

Could someone make a sample table of change for this particular equation, so i can see how it works.

Thanks, I will remember to select best answe

Choose values for theta in the interval (0, 2PI);
Since we have an r^2 (which is always positive), 36 cos(theta) is negative on (PI/2, 3PI/2)

Therefore, choose theta on [0, PI/2] and [ 3PI/2, 2PI]

compute r^2 for each value.
for example, when theta = PI/4, r^2 = 36 cos(PI/4)
r^2 = 36 /sqrt(2)
r^2 = 25.455844
r = -5.045 and 5.045
The points are (-5.045, PI/4) and (5.045, PI/4)

Here is a table of values:
choose theta = 0,pi/6,pi/4,pi/3,pi/2, 5pi/3,7pi/4,11pi/6, 2pi
Evaluate 36 cos(theta) for each theta
That is your r^2
Then, r has two solutions for each: one minus, one plus

Plot these values.