Question

5 multiple answer 20 points solve the following equation for ( 0 leq \theta leq 2 pi ). ( sin \theta = -\frac{1}{2} ) ( \frac{7 pi}{6} ) ( \frac{5 pi}{6} ) ( \frac{pi}{6} ) ( \frac{11 pi}{6} )

Explanation:

Step1: Recall the unit - circle values

We know that \(\sin\theta=\frac{1}{2}\) when \(\theta = \frac{\pi}{6}\) (in the first - quadrant) and \(\theta=\frac{5\pi}{6}\) (in the second - quadrant). Since \(\sin\theta=-\frac{1}{2}\), we need to find the angles in the third and fourth quadrants.

Step2: Find the angle in the third quadrant

For the third - quadrant, \(\theta=\pi+\frac{\pi}{6}\). Using the formula \(\theta=\pi + \alpha\) (where \(\alpha\) is the reference angle and \(\sin\alpha=\frac{1}{2}\), \(\alpha=\frac{\pi}{6}\)), we get \(\theta=\frac{7\pi}{6}\).

Step3: Find the angle in the fourth quadrant

For the fourth - quadrant, \(\theta = 2\pi-\frac{\pi}{6}\). Using the formula \(\theta = 2\pi-\alpha\) (where \(\alpha\) is the reference angle and \(\sin\alpha=\frac{1}{2}\), \(\alpha=\frac{\pi}{6}\)), we get \(\theta=\frac{11\pi}{6}\).

Answer:

\(\frac{7\pi}{6}\), \(\frac{11\pi}{6}\)