Question

Principal Solution of $ (5 \sin \theta)(2 \cos \theta + 1) = 0 $

• A. \( \frac{2\pi}{3} \), \( \frac{4\pi}{3} \)
• B. \( \frac{2\pi}{3} \), \( \frac{1}{3} \), \( \frac{2\pi}{3} \)
• C. \( \frac{3\pi}{3} \), \( \frac{1}{3} \), \( \frac{2\pi}{3} \)
• D. \( \frac{5\pi}{3} \), \( \frac{1}{3} \), \( \frac{2\pi}{3} \)

Hint:

When solving trigonometric equations, remember to break the equation into factors. Find the solutions for each factor individually.

Answer 1

The Correct Option is A

The equation \( (5 \sin \theta)(2 \cos \theta + 1) = 0 \) is a product of two factors. For the equation to be zero, one of the factors must be zero. This implies either \( 5 \sin \theta = 0 \) or \( 2 \cos \theta + 1 = 0 \).
- If \( 5 \sin \theta = 0 \), then \( \theta = n\pi \), where \( n \) is an integer.
- If \( 2 \cos \theta + 1 = 0 \), this simplifies to \( \cos \theta = -\frac{1}{2} \). The solutions for this are \( \theta = \frac{2\pi}{3} \) and \( \theta = \frac{4\pi}{3} \).
Therefore, the principal solutions are \( \frac{2\pi}{3} \) and \( \frac{4\pi}{3} \).