To solve the equation:

$2\sin(\theta) = -\sqrt{3}$

we’ll first isolate $\sin(\theta)$ by dividing both sides by 2:

$\sin(\theta) = -\frac{\sqrt{3}}{2}$

We need to find the values of $\theta$ for which $\sin(\theta) = -\frac{\sqrt{3}}{2}$ within the interval $0 \leq \theta < 2\pi$.

Step 1: Determine Reference Angles

The sine function equals $-\frac{\sqrt{3}}{2}$ at angles where the reference angle is $\frac{\pi}{3}$ (or $60^\circ$) in the third and fourth quadrants, where sine is negative.

Step 2: Find Solutions

The values of $\theta$ that satisfy $\sin(\theta) = -\frac{\sqrt{3}}{2}$ are:

1. Third quadrant: $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$
2. Fourth quadrant: $\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$

Final Answer

The solutions on the interval $0 \leq \theta < 2\pi$ are:

$\theta = \frac{4\pi}{3} \text{ and } \theta = \frac{5\pi}{3}$

Would you like more details on any steps?

---

Here are some related questions you might find useful:

1. What are the general solutions for $\sin(\theta) = -\frac{\sqrt{3}}{2}$?
2. How do you find reference angles for trigonometric functions?
3. What are the sine and cosine values of commonly used angles like $\frac{\pi}{3}$?
4. How can I solve similar equations involving cosine or tangent?
5. What strategies can help in solving trigonometric equations efficiently?

Tip: Remember that knowing the unit circle well can simplify solving trigonometric equations significantly.