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:

Use graphical or numerical methods to verify the count of solutions for trigonometric equations.

Answer 1

The Correct Option is B

To solve the equation \(\sin(9\theta) = \sin(\theta)\) within the interval \([0,2\pi]\), we use 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 \(\sin(9\theta) = \sin(\theta)\), we get two sets of equations:

\(9\theta = \theta + 2k\pi\)
\(9\theta = \pi - \theta + 2k\pi\)

Simplifying the first equation:

\(9\theta - \theta = 2k\pi\)

\(8\theta = 2k\pi\)

\(\theta = \frac{k\pi}{4}\)

For \(\theta\) in \([0,2\pi]\), we calculate 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 \(k = 8, \theta = 2\pi\).

These are 9 solutions.

Now, simplifying the second equation:

\(9\theta + \theta = \pi + 2k\pi\)

\(10\theta = \pi(1 + 2k)\)

\(\theta = \frac{(2k + 1)\pi}{10}\)

For \(\theta\) in \([0,2\pi]\), we find values of \(k\):

When \(k = 0, \theta = \frac{\pi}{10}\).

When \(k = 1, \theta = \frac{3\pi}{10}\).

When \(k = 2, \theta = \frac{\pi}{2}\). (Already accounted)

When \(k = 3, \theta = \frac{7\pi}{10}\).

When \(k = 4, \theta = \pi\). (Already accounted)

When \(k = 5, \theta = \frac{13\pi}{10}\).

When \(k = 6, \theta = \frac{3\pi}{2}\). (Already accounted)

When \(k = 7, \theta = \frac{17\pi}{10}\).

These are 8 additional solutions.

Thus, the total number of solutions is \(9 + 8 = 17\).

The correct answer is: 17.

Answer 2

We are given that: \[ \sin 90^\circ = \sin \theta \] which simplifies to: \[ \sin 90^\circ - \sin \theta = 0 \] We use the identity for the difference of sines: \[ \sin A - \sin B = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right) \] This leads to: \[ 2 \cos \left( \frac{90^\circ + \theta}{2} \right) \sin \left( \frac{90^\circ - \theta}{2} \right) = 0 \] Now we have two cases: \[ \sin C - \sin D = 2 \cos \left( \frac{C + D}{2} \right) \sin \left( \frac{C - D}{2} \right) \] This gives us: \[ 2 \cos 5\theta \sin 4\theta = 0 \] Thus, we have two conditions: \[ \cos 5\theta = 0 \quad {or} \quad \sin 4\theta = 0 \] For \( \cos 5\theta = 0 \): \[ 5\theta = (2n + 1) \frac{\pi}{2} \] \[ \theta = \frac{(2n + 1) \pi}{10} \] For \( \sin 4\theta = 0 \): \[ 4\theta = n\pi \] \[ \theta = \frac{n\pi}{4} \] Now, substituting \( n = 0, 1, 2, \dots \), we get the solutions for \( \theta \): \[ \theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}, \frac{11\pi}{10}, \frac{13\pi}{10}, \frac{15\pi}{10}, \frac{17\pi}{10}, \frac{19\pi}{10} \] For the second set of solutions: \[ \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{7\pi}{4}, 2\pi \] Thus, we have a total of 17 solutions, and the common solutions are \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \). 
Total number of solutions = 17