Find the points of intersection of the graphs of the equations. r = 1 + cos θ r = 1 − sin θ r ≥ 0, 0 ≤ θ < 2π

This question is based on the point of intersection.Therefore, the points of intersection of the graphs of the equations at 0 ≤ θ < 2π are:

Given:

Equations: r = 1 + cos θ                                                                      ...(1)

                  r = 1 − sin  θ                                                                      ...(2)                                                                

Where, r ≥ 0, 0 ≤ θ < 2π

We need to determined the point of intersection of the graphs of the equations.

To obtain the points of intersection, Equate the two equations above as follows;

r = 1 + cos θ = 1 - sin θ

=> 1 + cos θ = 1 - sin θ

Solve further for θ. We get,

1 + cos θ = 1 - sinθ

cos θ  = - sinθ

Now dividing both sides by - cos θ and solve it further,

To get the 2nd quadrant value of θ, add π ( = 180°) to the value of θ. i.e

Similarly, to get the fourth quadrant value of θ, add 2π ( = 360° ) to the value of θ. i.e

Therefore, the values of θ are 3π / 4 and 7π / 4.

Now substitute these values into equations (i) and (ii) as follows;

Represent the results  above in polar coordinates of the form (r, θ). i.e

Therefore, at the pole where r = 0, is also one of the points of intersection.  

Therefore, the points of intersection of the graphs of the equations at 0 ≤ θ < 2π are:

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i do not know this sorry : (

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get