Question

Number of solutions of equations \( \sin 9\theta = \sin \theta \) in the interval \( [0, 2\pi] \) is

• A. 16
• B. 17
• C. 18
• D. 15

Hint:

Alternatively, you can solve \( 9\theta = n\pi + (-1)^n \theta \). For even \( n \), \( 8\theta = 2k\pi \); for odd \( n \), \( 10\theta = (2k+1)\pi \). Counting these while removing duplicates at \( \pi/2 \) and \( 3\pi/2 \) will yield the same result.

Answer 1

The Correct Option is B

The given equation is \(\sin 9\theta = \sin \theta\). We need to find the number of solutions in the interval \([0, 2\pi]\).

Recall the identity for sine that states:

\(\sin A = \sin B \iff A = n\pi + (-1)^n B,\ \text{for integer } n\)

Applying this identity to \(\sin 9\theta = \sin \theta\), we get:

1. \(9\theta = \theta + 2k\pi\) or \(9\theta = (k\pi - \theta)\) where \(k\) is an integer.

Let's solve these equations separately:

1. From \(9\theta = \theta + 2k\pi\), we have:
   • \(8\theta = 2k\pi\)
   • \(\theta = \frac{k\pi}{4}\)
2. From \(9\theta = k\pi - \theta\), we have:
   • \(10\theta = k\pi\)
   • \(\theta = \frac{k\pi}{10}\)

Among these solutions, the one for \(k = 0\) is common in both sets (\(\theta = 0\)), leading to overlapping. Therefore, the total number of distinct solutions is:

\(8 + 10 - 1 = 17\)

Thus, the number of solutions is 17.