find all solutions to ( 2 sin ( \theta ) = - sqrt { 3 } ) on the interval ( 0 leq \theta < 2 pi )

Explanation:

Step1: Solve for $\sin(\theta)$

Divide both sides of the equation (2\sin(\theta)=-\sqrt{3}) by (2). (\sin(\theta)=-\frac{\sqrt{3}}{2})

Step2: Find the reference angle

We know that (\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}). The reference angle (\alpha=\frac{\pi}{3})

Step3: Determine the quadrants

Since (\sin(\theta)<0), (\theta) lies in the third and fourth quadrants. In the third quadrant: (\theta=\pi+\alpha=\pi + \frac{\pi}{3}=\frac{4\pi}{3}) In the fourth quadrant: (\theta = 2\pi-\alpha=2\pi-\frac{\pi}{3}=\frac{5\pi}{3})

Answer:

(\theta=\frac{4\pi}{3},\frac{5\pi}{3})