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Unformatted text preview: or r = −r . If r = r , then θ = θ + 2πk , so one possibility is that an intersection point
P has a representation (r, θ) which satisﬁes r = 3 and another representation (r, θ + 2πk ) for
some integer, k which satisﬁes r = 6 cos(2θ). At this point, we replace every occurrence θ in
the equation r = 6 cos(2θ) with (θ +2πk ) and then see if, by equating the resulting expressions
for r, we get any more solutions for θ.13 Since cos(2(θ + 2πk )) = cos(2θ + 4πk ) = cos(2θ) for
every integer k , however, the equation r = 6 cos(2(θ + 2πk )) reduces to the same equation
we had before, r = 6 cos(2θ), which means we get no additional solutions. Moving on to
the case where r = −r , we have that θ = θ + (2k + 1)π for integers k . We look to see
if we can ﬁnd points P which have a representation (r, θ) that satisﬁes r = 3 and another,
12 See Example 11.5.2 number 3.
The authors have chosen to replace θ with θ + 2πk in the equation r = 6 cos(2θ) for illustration purposes only. We
could have just...
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