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**Unformatted text preview: **ouraged to check these
results algebraically and geometrically. 3. Proceeding as above, we ﬁrst graph r = 3 and r = 6 cos(2θ) to get an idea of how many
intersection points to expect and where they lie. The graph of r = 3 is a circle centered at
the origin with a radius of 3 and the graph of r = 6 cos(2θ) is another four-leafed rose.12
y
6 3 −6 −3 3 6 x −3 −6 r = 3 and r = 6 cos(2θ )
It appears as if there are eight points of intersection - two in each quadrant. We ﬁrst look to
see if there any points P (r, θ) with a representation that satisﬁes both r = 3 and r = 6 cos(2θ).
π
For these points, 6 cos(2θ) = 3 or cos(2θ) = 1 . Solving, we get θ = π + πk or θ = 56 + πk
2
6
for integers k . Out of all of these solutions, we obtain just four distinct points represented
π
π
by 3, π , 3, 56 , 3, 76 and 3, 11π . To determine the coordinates of the remaining four
6
6
points, we have to consider how the representations of the points of intersection can diﬀer. We
know from Section 11.4 that if (r, θ) and (r , θ ) represent the same point and r = 0, then either
r = r...

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