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

To solve the equation $\cos\theta + 1 = \sqrt{3} \sin\theta$, we first rewrite it as:
$$\sqrt{3} \sin\theta - \cos\theta = 1$$
Divide the entire equation by 2 to bring it into a standard trigonometric form:
$$\frac{\sqrt{3}}{2} \sin\theta - \frac{1}{2} \cos\theta = \frac{1}{2}$$
Recognizing that $\cos(\pi/6) = \sqrt{3}/2$ and $\sin(\pi/6) = 1/2$, we can use the identity $\sin A \cos B - \cos A \sin B = \sin(A - B)$:
$$\sin\theta \cos(\pi/6) - \cos\theta \sin(\pi/6) = \frac{1}{2}$$
$$\sin(\theta - \pi/6) = \frac{1}{2}$$
Now we find all values for $(\theta - \pi/6)$ in the interval corresponding to $\theta \in (-2\pi, 2\pi)$. The range for $x = \theta - \pi/6$ is $(-2\pi - \pi/6, 2\pi - \pi/6)$, which is $(-13\pi/6, 11\pi/6)$.
The values of $x$ where $\sin x = 1/2$ are:
$$x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{\pi}{6}-2\pi, \frac{5\pi}{6}-2\pi$$
$$x = \frac{\pi}{6}, \frac{5\pi}{6}, -\frac{11\pi}{6}, -\frac{7\pi}{6}$$
Solving for $\theta = x + \pi/6$:
1. $\theta = \pi/6 + \pi/6 = \pi/3$
2. $\theta = 5\pi/6 + \pi/6 = \pi$
3. $\theta = -11\pi/6 + \pi/6 = -10\pi/6 = -5\pi/3$
4. $\theta = -7\pi/6 + \pi/6 = -6\pi/6 = -\pi$
All these values lie within $(-2\pi, 2\pi)$. The sum of all elements in $S$ is:
$$\sum_{\theta \in S} \theta = \frac{\pi}{3} + \pi - \frac{5\pi}{3} - \pi = \frac{\pi - 5\pi}{3} = -\frac{4\pi}{3}$$