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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
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).
For these points, 6 cos(2θ) = 3 or cos(2θ) = 1 . Solving, we get θ = π + πk or θ = 56 + πk
for integers k . Out of all of these solutions, we obtain just four distinct points represented
by 3, π , 3, 56 , 3, 76 and 3, 11π . To determine the coordinates of the remaining four
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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