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**Unformatted text preview: **as easily chosen to do this substitution in the equation r = 3. Since there is no θ in r = 3, however,
this case would reduce to the previous case instantly. The reader is encouraged to follow this latter procedure in the
interests of eﬃciency.
13 812 Applications of Trigonometry
(−r, θ + (2k + 1)π ), that satisﬁes r = 6 cos(2θ). To do this, we substitute14 (−r) for r and
(θ + (2k + 1)π ) for θ in the equation r = 6 cos(2θ) and get −r = 6 cos(2(θ + (2k + 1)π )).
Since cos(2(θ + (2k + 1)π )) = cos(2θ + (2k + 1)(2π )) = cos(2θ) for all integers k , the equation
−r = 6 cos(2(θ + (2k + 1)π )) reduces to −r = 6 cos(2θ), or r = −6 cos(2θ). Coupling this
1
π
equation with r = 3 gives −6 cos(2θ) = 3 or cos(2θ) = − 2 . We get θ = π + πk or θ = 23 + πk .
3
From these solutions, we obtain15 the remaining four intersection points with representations
π
π
π
−3, π , −3, 23 , −3, 43 and −3, 53 , which we can readily check graphically.
3 θ
θ
4. As usual, we begin by graphing r = 3 sin 2 and...

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