Find all solutions to the equation.
$\sin\theta=-\frac{\sqrt3}{2}$

• A. $\theta=\frac{\pi}{3}+2\pi n,\frac{2\pi}{3}+2\pi n$
• B. $\theta=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n$
• C. $\theta=\frac{4\pi}{3}+2\pi n,\frac{5\pi}{3}+2\pi n$
• D. $\theta=\frac{\pi}{6}+2\pi n,\frac{5\pi}{6}+2\pi n$

Answer: (C)

Verified step by step guidance

1. Identify the given equation: \( \sin \theta = -\frac{\sqrt{3}}{2} \). This indicates that we are looking for angles where the sine value is \(-\frac{\sqrt{3}}{2}\).
2. Recall that the sine function is negative in the third and fourth quadrants. The reference angle for \( \sin \theta = \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{3} \).
3. Determine the angles in the third and fourth quadrants that have the same reference angle. These are \( \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \) and \( \theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \).
4. Express the general solutions for these angles by adding multiples of the full period of the sine function, which is \( 2\pi \). Thus, the solutions are \( \theta = \frac{4\pi}{3} + 2\pi n \) and \( \theta = \frac{5\pi}{3} + 2\pi n \), where \( n \) is an integer.
5. Verify the solutions by checking that they satisfy the original equation \( \sin \theta = -\frac{\sqrt{3}}{2} \) and are consistent with the sine function's periodicity and quadrant rules.