Question

If the general solution set of \( \sin x + 3 \sin 3x + \sin 5x = 0 \) is \( S \), then \[ \sin a \quad {for} \quad a \in S \quad {is} \quad \{ \sin a \mid a \in S \} = \]

• A. \( \{1, -1, 0\} \)
• B. \( \left\{ \frac{1}{2}, -\frac{1}{2}, 0, 1, -1 \right\} \)
• C. \( \left\{ \frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2} \right\} \)
• D. \( \left\{ 1, -1, \frac{\sqrt{3}}{2}, 0, -\frac{\sqrt{3}}{2} \right\} \)

Hint:

When solving trigonometric equations involving multiple sine terms, use trigonometric identities to simplify and solve for possible values.

Answer 1

The Correct Option is C

We are given the equation: \[ \sin x + 3 \sin 3x + \sin 5x = 0. \] The general solution set of this equation involves finding values of \( x \) that satisfy it. 
Step 1: We use the known identity for the sum of sines: \[ \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right). \] Applying this identity to the terms \( \sin x + \sin 5x \), we get: \[ \sin x + \sin 5x = 2 \sin \left( \frac{x + 5x}{2} \right) \cos \left( \frac{5x - x}{2} \right) = 2 \sin(3x) \cos(2x). \]
Step 2: Now, the equation becomes: \[ 2 \sin(3x) \cos(2x) + 3 \sin(3x) = 0. \] Factor out \( \sin(3x) \): \[ \sin(3x) \left( 2 \cos(2x) + 3 \right) = 0. \]
Step 3: This gives us two cases to solve: 1. \( \sin(3x) = 0 \), 2. \( 2 \cos(2x) + 3 = 0 \). For the first case, \( \sin(3x) = 0 \), the solutions are: \[ x = \frac{n\pi}{3}, \quad n \in \mathbb{Z}. \] For the second case, \( \cos(2x) = -\frac{3}{2} \), which has no real solutions. 
Step 4: Thus, the general solutions for \( x \) are of the form \( x = \frac{n\pi}{3}, \quad n \in \mathbb{Z} \).
Now, we calculate the possible values of \( \sin a \) for these solutions. Since the sine function takes values between \( -1 \) and \( 1 \), the possible values of \( \sin a \) are: \[ \boxed{ \left\{ \frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2} \right\} }. \]