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. Start by understanding that the equation \( \sin\theta = -\frac{\sqrt{3}}{2} \) implies that \( \theta \) is in the third or fourth quadrant, where the sine function is negative.
2. Recall the reference angle for \( \sin\theta = \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{3} \). This means the angles in the unit circle where sine is \( -\frac{\sqrt{3}}{2} \) are \( \theta = \pi + \frac{\pi}{3} \) and \( \theta = 2\pi - \frac{\pi}{3} \).
3. Calculate these angles: \( \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \) and \( \theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \).
4. Since the sine function is periodic with period \( 2\pi \), the general solutions for \( \theta \) 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 substituting back into the original equation to ensure they satisfy \( \sin\theta = -\frac{\sqrt{3}}{2} \).