To find the zeros and the maximum value of the function \( r = 3 - 6 \sin \theta \), we need to analyze the trigonometric function involved.
### Finding the Zeros
The zeros of the function occur where \( r = 0 \). Therefore, we set the function equal to zero and solve for \( \theta \):
[tex]\[ 3 - 6 \sin \theta = 0 \][/tex]
Isolate \(\sin \theta\):
[tex]\[ 6 \sin \theta = 3 \][/tex]
[tex]\[ \sin \theta = \frac{1}{2} \][/tex]
We know from basic trigonometry that \(\sin \theta = \frac{1}{2}\) at two points in the range \( 0 \leq \theta < 2\pi \):
[tex]\[ \theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \frac{5\pi}{6} \][/tex]
Thus, the zeros of the function \( r = 3 - 6 \sin \theta \) are at \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \).
### Finding the Maximum Value
To find the maximum value of the function, we need to determine when \( r = 3 - 6 \sin \theta \) reaches its highest point. The sine function, \(\sin \theta\), takes values between \(-1\) and \(1\). Therefore, the minimum and maximum values of \( 6 \sin \theta \) are \(-6\) and \(6\), respectively.
We substitute these extremum values into the function \( r = 3 - 6 \sin \theta \):
1. When \(\sin \theta = -1\):
[tex]\[ r = 3 - 6(-1) = 3 + 6 = 9 \][/tex]
2. When \(\sin \theta = 1\):
[tex]\[ r = 3 - 6(1) = 3 - 6 = -3 \][/tex]
Clearly, the function \( r \) reaches its maximum value when \(\sin \theta = -1\). Therefore, the maximum value of \( r = 3 - 6 \sin \theta \) is:
[tex]\[ r = 9 \][/tex]
### Summary
- The zeros of \( r = 3 - 6 \sin \theta \) are at \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \).
- The maximum value of [tex]\( r = 3 - 6 \sin \theta \)[/tex] is [tex]\( r = 9 \)[/tex].
