Question

Let \( S = \{\theta \in (-2\pi, 2\pi): \cos \theta + 1 = \sqrt{3} \sin \theta\} \). Then \( \sum_{\theta \in S} \theta \) is equal to:

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

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The equation is of the form \( a \cos \theta + b \sin \theta = c \). We divide by \( \sqrt{a^2 + b^2} \) to transform it into a single trigonometric ratio.
Step 2: Key Formula or Approach:
1. \( \sqrt{3} \sin \theta - \cos \theta = 1 \) 2. Divide by \( \sqrt{(\sqrt{3})^2 + (-1)^2} = 2 \): \( \frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta = \frac{1}{2} \implies \sin(\theta - \frac{\pi}{6}) = \frac{1}{2} \)
Step 3: Detailed Explanation:
1. \( \sin(\theta - \pi/6) = \sin(\pi/6) \)
2. General solution: \( \theta - \pi/6 = n\pi + (-1)^n \frac{\pi}{6} \)
3. For \( n = 0 \): \( \theta = \pi/3 \)
4. For \( n = 1 \): \( \theta = \pi \)
5. For \( n = -1 \): \( \theta = -2\pi/3 + \pi/6 = -\pi \) (Note: \( \pi \) and \( -\pi \) are in the range)
6. For \( n = -2 \): \( \theta = -2\pi + \pi/6 + \pi/6 = -5\pi/3 \)
7. Sum of values in \( (-2\pi, 2\pi) \): \( \frac{\pi}{3} + \pi - \pi - \frac{5\pi}{3} = -\frac{4\pi}{3} \)
Step 4: Final Answer:
The sum of the values of \( \theta \) is \( -\frac{4\pi}{3} \).