How do you graph $r=4+6costheta$ ?

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Answer 1 · 2016-11-15T07:59:28

For making this cardioid-like graph, see explanation. The graph that is inserted is for the equivalent cartesian form $x^2+y^2-6x-4sqrt(x^2+y^2)=0$

$r=4+6 cos theta>=0 to cos theta >=-2/3 to$

$theta in (-131.81^o, 131.81^o)$ , nearly.

As $cos(-theta)=costheta$ , the graph is symmetrical with respect to

the initial line $theta = 0$ .

The period of $r(theta)$ is $2pi$ .

So, a Table for $theta in (1. 131.61^o)$ is sufficient, for making the

graph..

$(r, theta): (10, 0^o) (7, 60^o) (4, 90^o) (1, 120^o) (0, 131.81^o)$

The other half can be made using symmetry about the axis

$theta=0$ .
graph{x^2 + y^2- 6x - 4sqrt (x^2+y^2) =0 [-10, 10, -5, 5]}

How do you graph r=4+6costhetar=4+6costheta?