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

Step 1: Understanding the Concept:
To find the number of solutions, we use the general solution for the trigonometric equation \( \sin \alpha = \sin \beta \), which is \( \alpha = n\pi + (-1)^n \beta \), or more simply, we analyze the two primary cases: \( \alpha - \beta = 2k\pi \) and \( \alpha + \beta = (2k+1)\pi \).

Step 2: Key Formula or Approach:
The equation \( \sin 9\theta - \sin \theta = 0 \) can be factored using the sum-to-product identity: \[ 2 \cos\left(\frac{9\theta + \theta}{2}\right) \sin\left(\frac{9\theta - \theta}{2}\right) = 0 \] \[ 2 \cos(5\theta) \sin(4\theta) = 0 \]

Step 3: Detailed Explanation:
This gives us two cases: Case 1: \( \cos(5\theta) = 0 \) \[ 5\theta = (2k+1)\frac{\pi}{2} \implies \theta = \frac{(2k+1)\pi}{10} \] For \( \theta \in [0, 2\pi] \), \( k \) can be \( 0, 1, 2, \dots, 9 \). Total solutions in Case 1 = 10. Case 2: \( \sin(4\theta) = 0 \) \[ 4\theta = k\pi \implies \theta = \frac{k\pi}{4} \] For \( \theta \in [0, 2\pi] \), \( k \) can be \( 0, 1, 2, \dots, 8 \). Total solutions in Case 2 = 9. Overlapping Solutions: Check where \( \frac{(2k_1+1)\pi}{10} = \frac{k_2\pi}{4} \). \[ 4(2k_1+1) = 10k_2 \implies 2(2k_1+1) = 5k_2 \] Since the left side is even but not divisible by 4, and contains a factor of 2, \( k_2 \) must be an even multiple of 2 (specifically, \( k_2 = 2, 6 \)). - If \( k_2 = 2 \), \( \theta = \pi/2 \). - If \( k_2 = 6 \), \( \theta = 3\pi/2 \). There are 2 overlapping solutions. Total unique solutions = \( 10 + 9 - 2 = 17 \).

Step 4: Final Answer
The number of solutions in the interval \( [0, 2\pi] \) is 17.