To resolve the equation \(\sin(9\theta) = \sin(\theta)\) within the interval \([0,2\pi]\), we apply the identity: \(\sin A = \sin B\) implies \(A = B + 2k\pi\) or \(A = \pi - B + 2k\pi\), where \(k \in \mathbb{Z}\).
Applying this to \(\sin(9\theta) = \sin(\theta)\) yields two sets of equations:
\(9\theta = \theta + 2k\pi\)
\(9\theta = \pi - \theta + 2k\pi\)
The first equation simplifies to:
\(8\theta = 2k\pi\)
\(\theta = \frac{k\pi}{4}\)
For \(\theta\) in \([0,2\pi]\), we determine the values of \(k\):
When \(k = 0, \theta = 0\).
When \(k = 1, \theta = \frac{\pi}{4}\).
When \(k = 2, \theta = \frac{\pi}{2}\).
When \(k = 3, \theta = \frac{3\pi}{4}\).
When \(k = 4, \theta = \pi\).
When \(k = 5, \theta = \frac{5\pi}{4}\).
When \(k = 6, \theta = \frac{3\pi}{2}\).
When \(k = 7, \theta = \frac{7\pi}{4}\).
When \(k = 8, \theta = 2\pi\).
This provides 9 solutions.
Next, the second equation is simplified:
\(10\theta = \pi + 2k\pi\)
\(10\theta = \pi(1 + 2k)\)
\(\theta = \frac{(2k + 1)\pi}{10}\)
For \(\theta\) in \([0,2\pi]\), we find the values of \(k\):
When \(k = 0, \theta = \frac{\pi}{10}\).
When \(k = 1, \theta = \frac{3\pi}{10}\).
When \(k = 2, \theta = \frac{5\pi}{10} = \frac{\pi}{2}\). (Duplicate)
When \(k = 3, \theta = \frac{7\pi}{10}\).
When \(k = 4, \theta = \frac{9\pi}{10}\).
When \(k = 5, \theta = \frac{11\pi}{10}\).
When \(k = 6, \theta = \frac{13\pi}{10}\).
When \(k = 7, \theta = \frac{15\pi}{10} = \frac{3\pi}{2}\). (Duplicate)
When \(k = 8, \theta = \frac{17\pi}{10}\).
When \(k = 9, \theta = \frac{19\pi}{10}\).
The values that are duplicates from the first set are \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\). Thus, there are \(10 - 2 = 8\) new unique solutions.
The total number of solutions is \(9 + 8 = 17\).
The final answer is: 17.