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To find the intersection points of the curves defined by the polar equations [tex]\( r = 5 + 4\sin(\theta) \)[/tex] and [tex]\( r = -6\sin(\theta) \)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex], we need to check if the given candidate points satisfy both equations simultaneously.
### Candidate Points:
1. [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex]
2. [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex]
3. [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex]
4. [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
5. [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
We will check each point one by one.
#### Check 1: [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since [tex]\((-3) \neq 3\)[/tex], [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex] is NOT an intersection point.
#### Check 2: [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since [tex]\((-3) \neq 3\)[/tex], [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex] is NOT an intersection point.
#### Check 3: [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{5\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(\frac{1}{2}\right) = 5 + 2 = 7 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{5\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(\frac{1}{2}\right) = -3 \)[/tex].
Since [tex]\(3 \neq 7\)[/tex] and [tex]\(3 \neq -3\)[/tex], [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex] is NOT an intersection point.
#### Check 4: [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since both equations are satisfied ([tex]\(3 = 3\)[/tex]), [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex] is an intersection point.
#### Check 5: [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since both equations are satisfied ([tex]\(3 = 3\)[/tex]), [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex] is an intersection point.
### Conclusion:
The intersection points of the polar equations [tex]\( r = 5 + 4\sin(\theta) \)[/tex] and [tex]\( r = -6\sin(\theta) \)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex] are:
- [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
- [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
Thus, the correct points are:
[tex]\[ \left(3, \frac{7\pi}{6}\right), \left(3, \frac{11\pi}{6}\right) \][/tex]
### Candidate Points:
1. [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex]
2. [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex]
3. [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex]
4. [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
5. [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
We will check each point one by one.
#### Check 1: [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since [tex]\((-3) \neq 3\)[/tex], [tex]\(\left(-3, \frac{7\pi}{6}\right)\)[/tex] is NOT an intersection point.
#### Check 2: [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since [tex]\((-3) \neq 3\)[/tex], [tex]\(\left(-3, \frac{11\pi}{6}\right)\)[/tex] is NOT an intersection point.
#### Check 3: [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{5\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(\frac{1}{2}\right) = 5 + 2 = 7 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{5\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(\frac{1}{2}\right) = -3 \)[/tex].
Since [tex]\(3 \neq 7\)[/tex] and [tex]\(3 \neq -3\)[/tex], [tex]\(\left(3, \frac{5\pi}{6}\right)\)[/tex] is NOT an intersection point.
#### Check 4: [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{7\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since both equations are satisfied ([tex]\(3 = 3\)[/tex]), [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex] is an intersection point.
#### Check 5: [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
- Calculate [tex]\(r_1\)[/tex]: [tex]\( r = 5 + 4\sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = 5 + 4\left(-\frac{1}{2}\right) = 5 - 2 = 3 \)[/tex].
- Calculate [tex]\(r_2\)[/tex]: [tex]\( r = -6 \sin\left(\frac{11\pi}{6}\right) \)[/tex].
- [tex]\(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\)[/tex]
- [tex]\( r = -6\left(-\frac{1}{2}\right) = 3 \)[/tex].
Since both equations are satisfied ([tex]\(3 = 3\)[/tex]), [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex] is an intersection point.
### Conclusion:
The intersection points of the polar equations [tex]\( r = 5 + 4\sin(\theta) \)[/tex] and [tex]\( r = -6\sin(\theta) \)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex] are:
- [tex]\(\left(3, \frac{7\pi}{6}\right)\)[/tex]
- [tex]\(\left(3, \frac{11\pi}{6}\right)\)[/tex]
Thus, the correct points are:
[tex]\[ \left(3, \frac{7\pi}{6}\right), \left(3, \frac{11\pi}{6}\right) \][/tex]
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