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. $\frac{2\pi}{3}$
• B. $\frac{4\pi}{3}$
• C. $-\frac{2\pi}{3}$
• D. $-\frac{4\pi}{3}$

Hint:

Transform the equation into the form $\sin(\theta - \phi) = k$ or use half-angle substitutions to find all valid roots in the specified domain.

Answer 1

The Correct Option is D

Given equation: $1 + \cos\theta = \sqrt{3} \sin\theta$.
Using half-angle formulas $1 + \cos\theta = 2\cos^2(\theta/2)$ and $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$, we get:
$$2\cos^2(\theta/2) = \sqrt{3} \cdot 2\sin(\theta/2)\cos(\theta/2)$$
$$2\cos(\theta/2) [\cos(\theta/2) - \sqrt{3}\sin(\theta/2)] = 0$$
This gives two possibilities:
Case 1: $\cos(\theta/2) = 0 \implies \theta/2 = n\pi + \pi/2 \implies \theta = 2n\pi + \pi$.
In the range $(-2\pi, 2\pi)$, the solutions are $\theta = \pi$ (for $n=0$) and $\theta = -\pi$ (for $n=-1$).
Case 2: $\cos(\theta/2) - \sqrt{3}\sin(\theta/2) = 0 \implies \tan(\theta/2) = 1/\sqrt{3}$.
The general solution is $\theta/2 = n\pi + \pi/6 \implies \theta = 2n\pi + \pi/3$.
In the range $(-2\pi, 2\pi)$, the solutions are $\theta = \pi/3$ (for $n=0$) and $\theta = -2\pi + \pi/3 = -5\pi/3$ (for $n=-1$).
The set of solutions is $S = \{-\pi, \pi, \pi/3, -5\pi/3\}$.
Summing the elements: $\sum_{\theta \in S} \theta = -\pi + \pi + \pi/3 - 5\pi/3 = -4\pi/3$.