# How do you graph #r=2cos3theta#?

See Socratic graph for the Cartesian form of the equation

So, for one loop of this 3-petal rose,

For the other half of one period r is negative and, if you choose to

admit negative r, the third loop would be redrawn.

I have used the Cartesian form of the equation, using

graph{(x^2+y^2)^2+6x(x^2+y^2)=8x^3 [-4, 4, -2, 2]}

I introduce here a 2-cosine-combined 10-petal rose for

my viewers of this answer.

Observe that four of the petals are much smaller.

graph{0.01((x^2+y^2)^2.5-x^4+6x^2y^2-y^4)(0.25(x^2+y^2)^3.5-x^6+15x^4y^2-15x^2y^4+y^6)(x^2+y^2-.04)=0}

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To graph ( r = 2\cos(3\theta) ), follow these steps:

- Plot points for various values of ( \theta ) ranging from ( 0 ) to ( 2\pi ).
- Substitute each value of ( \theta ) into the equation ( r = 2\cos(3\theta) ) to find the corresponding value of ( r ).
- Use polar coordinates to plot each point ( (r, \theta) ) on the graph.
- Connect the points to form the graph of the equation.

Keep in mind that ( r ) represents the distance from the origin (the pole), and ( \theta ) represents the angle measured counterclockwise from the positive x-axis.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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