How do you graph #r=3sintheta+3#?

Question
Answer 1

#r = a ( 1+cos (theta-alpha))#.

#alpha# is the second parameter. The line #theta=alpha# gives

The whole cardioid can be traced for one period #2pi#, for example

#theta in (alpha, alpha+2pi)#,, .

Here,# r = 3(1+ cos (theta-pi/2)#, and so, #a = 3 and theta=alpha =

The half of this cardioid in #Q_4 and Q_1# can be traced, using

#(r, theta): (0, 3/2pi) (3-13sqrt2, 7/4pi) (3, 0) (3+3/sqrt2, pi/4) (6, pi/2)#.

The other half in #Q_2 and Q_3# can be drawn as mirror image,

with respect to the line of symmetry, #theta = pi/2#.

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