Question

The number of solutions of the equation \[ \sin 7\theta - \sin 3\theta = \sin 4\theta \] that lie in the interval \( (0, \pi) \) is:

• A. \(6\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

Use trigonometric identities to simplify equations before solving for roots in a given interval.

Answer 1

The Correct Option is D

Using the identity \(\sin C - \sin D = 2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)\), we have: \[ 2\cos\left(\frac{7\theta + 3\theta}{2}\right)\sin\left(\frac{7\theta - 3\theta}{2}\right) = \sin 4\theta \] \[ 2\cos(5\theta)\sin(2\theta) = \sin 4\theta \] \[ 2\cos(5\theta)\sin(2\theta) = 2\sin(2\theta)\cos(2\theta) \] \[ 2\sin(2\theta)\left(\cos(5\theta) - \cos(2\theta)\right) = 0 \] So, either \(\sin(2\theta) = 0\) or \(\cos(5\theta) - \cos(2\theta) = 0\). Case 1: \(\sin(2\theta) = 0\) \[2\theta = n\pi\] \[\theta = \frac{n\pi}{2}\] Since \(0<\theta<\pi\), we have \(0<\frac{n\pi}{2}<\pi\), so \(0<n<2\). Thus, \(n = 1\), and \(\theta = \frac{\pi}{2}\). Case 2: \(\cos(5\theta) - \cos(2\theta) = 0\) \[\cos(5\theta) = \cos(2\theta)\] \[5\theta = 2n\pi \pm 2\theta\] Subcase 2.1: \(5\theta = 2n\pi + 2\theta\) \[3\theta = 2n\pi\] \[\theta = \frac{2n\pi}{3}\] Since \(0<\theta<\pi\), we have \(0<\frac{2n\pi}{3}<\pi\), so \(0<2n<3\), which means \(0<n<\frac{3}{2}\). Thus, \(n = 1\), and \(\theta = \frac{2\pi}{3}\). Subcase 2.2: \(5\theta = 2n\pi - 2\theta\) \[7\theta = 2n\pi\] \[\theta = \frac{2n\pi}{7}\] Since \(0<\theta<\pi\), we have \(0<\frac{2n\pi}{7}<\pi\), so \(0<2n<7\), which means \(0<n<\frac{7}{2}\). Thus, \(n = 1, 2, 3\), and \(\theta = \frac{2\pi}{7}, \frac{4\pi}{7}, \frac{6\pi}{7}\). The solutions are \(\frac{\pi}{2}, \frac{2\pi}{3}, \frac{2\pi}{7}, \frac{4\pi}{7}, \frac{6\pi}{7}\). There are 5 solutions in the interval \((0, \pi)\). Final Answer: The final answer is $\boxed{(4)}$