To solve for all angles $\theta$ in the interval $0^\circ \leq \theta < 360^\circ$ where

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

we proceed as follows:

Step 1: Analyze the sine function

The sine function is negative in Quadrants III and IV of the unit circle because sine corresponds to the vertical coordinate (negative below the $x$-axis).

The reference angle where $\sin \theta = \frac{\sqrt{3}}{2}$ is $60^\circ$. This is based on standard trigonometric values.

Step 2: Determine the solutions in Quadrants III and IV

• In Quadrant III, the angle is: $\theta = 180^\circ + \text{reference angle} = 180^\circ + 60^\circ = 240^\circ.$

• In Quadrant IV, the angle is: $\theta = 360^\circ - \text{reference angle} = 360^\circ - 60^\circ = 300^\circ.$

Step 3: Verify and summarize

The two solutions in the interval $0^\circ \leq \theta < 360^\circ$ are:

$\boxed{\theta = 240^\circ \text{ and } 300^\circ.}$

Additional Check:

• For $\theta = 240^\circ$: $\sin(240^\circ) = -\frac{\sqrt{3}}{2}.$
• For $\theta = 300^\circ$: $\sin(300^\circ) = -\frac{\sqrt{3}}{2}.$

Both solutions satisfy the equation.

Let me know if you need further clarification or a graphical explanation! Here are 5 related questions for practice:

1. Solve $\sin \theta = \frac{\sqrt{3}}{2}$ for $0^\circ \leq \theta < 360^\circ$.
2. Solve $\cos \theta = -\frac{1}{2}$ for $0^\circ \leq \theta < 360^\circ$.
3. Solve $\tan \theta = \sqrt{3}$ for $0^\circ \leq \theta < 360^\circ$.
4. Find the general solution for $\sin \theta = -\frac{\sqrt{3}}{2}$ in degrees.
5. Solve $\sin \theta = -0.5$ for $0^\circ \leq \theta < 360^\circ$.

Tip: Always identify the reference angle and check which quadrants the trigonometric function is positive or negative.