Solve
sin ( θ )

To solve the equation $\sin(\theta) = -\frac{\sqrt{3}}{2}$ on the interval $[0, 2\pi]$, follow these steps:

Step 1: Identify the reference angle

The sine of an angle equals $-\frac{\sqrt{3}}{2}$ when the reference angle $\alpha$ is $\frac{\pi}{3}$. This reference angle corresponds to the angle where $\sin(\alpha) = \frac{\sqrt{3}}{2}$, but we want the negative value, so we'll look in the appropriate quadrants.

Step 2: Determine the quadrants

Since the sine function is negative, $\theta$ must be in the third or fourth quadrants:

• In the third quadrant: $\theta = \pi + \frac{\pi}{3}$
• In the fourth quadrant: $\theta = 2\pi - \frac{\pi}{3}$

Step 3: Calculate the specific angles

• In the third quadrant: $\theta = \pi + \frac{\pi}{3} = \frac{3\pi}{3} + \frac{\pi}{3} = \frac{4\pi}{3}$
• In the fourth quadrant: $\theta = 2\pi - \frac{\pi}{3} = \frac{6\pi}{3} - \frac{\pi}{3} = \frac{5\pi}{3}$

Step 4: Provide the solution set

The solutions on the interval $[0, 2\pi]$ are: $\theta = \frac{4\pi}{3}, \frac{5\pi}{3}$

Would you like more details or have any questions?

Related Questions:

1. What are the values of $\sin(\theta)$ in all four quadrants?
2. How do you determine the reference angle for a given sine value?
3. Can you explain how the unit circle is used to find these solutions?
4. What is the general solution for $\sin(\theta) = -\frac{\sqrt{3}}{2}$?
5. How does the periodicity of sine affect the solutions?

Tip:

Always check which quadrants correspond to positive or negative sine values to determine the correct angles.