To solve the equation $2\sin(\theta) = -\sqrt{3}$ on the interval $0 \leq \theta < 2\pi$, let’s proceed step by step:

Step 1: Isolate $\sin(\theta)$

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

Step 2: Identify reference angle

The reference angle for $\sin(\theta) = \frac{\sqrt{3}}{2}$ is $\frac{\pi}{3}$ (from the unit circle).

Step 3: Determine quadrants where $\sin(\theta)$ is negative

The sine function is negative in the third and fourth quadrants.

Thus, the solutions are in:

• Quadrant III: $\pi + \text{reference angle} = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$
• Quadrant IV: $2\pi - \text{reference angle} = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$

Step 4: Write final solutions

The solutions to the equation $2\sin(\theta) = -\sqrt{3}$ on the interval $0 \leq \theta < 2\pi$ are: $\theta = \frac{4\pi}{3}, \frac{5\pi}{3}$

If you need further clarification or want additional details, let me know!

Follow-up questions to deepen understanding:

1. How can you verify these solutions using a calculator or unit circle?
2. Why does the sine function have negative values in specific quadrants?
3. What is the importance of the interval $0 \leq \theta < 2\pi$ in trigonometric equations?
4. How would the solutions differ if the interval was $-2\pi \leq \theta < 0$?
5. What are the solutions to $2\sin(\theta) = \sqrt{3}$ instead?

Tip:

Always sketch the unit circle when solving trigonometric equations. It helps visualize angles and quadrants more effectively!