solve the following equation for 0<=theta<=2pi: sin(2*theta) =-1/2

To solve the equation \( \sin(2\theta) = -\frac{1}{2} \) for \( 0 \leq \theta \leq 2\pi \), we start by finding the values of \( 2\theta \) that satisfy the equation, and then we solve for \( \theta \).

### Step 1: Solve for \( 2\theta \)

The equation is:

\[
\sin(2\theta) = -\frac{1}{2}
\]

The sine function equals \(-\frac{1}{2}\) at specific angles within its period. The general solutions for \( x \) where \( \sin(x) = -\frac{1}{2} \) are:

\[
x = \frac{7\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{11\pi}{6} + 2k\pi, \quad \text{for any integer } k.
\]

### Step 2: Apply these solutions to \( 2\theta \)

Since \( x = 2\theta \), we have:

\[
2\theta = \frac{7\pi}{6} + 2k\pi \quad \text{or} \quad 2\theta = \frac{11\pi}{6} + 2k\pi.
\]

Now, solve for \( \theta \):

\[
\theta = \frac{7\pi}{12} + k\pi \quad \text{or} \quad \theta = \frac{11\pi}{12} + k\pi.
\]

### Step 3: Find values of \( \theta \) in the range \( 0 \leq \theta \leq 2\pi \)

We will find the values of \( \theta \) by setting \( k = 0 \) and \( k = 1 \):

1. For \( k = 0 \):
   \[
   \theta = \frac{7\pi}{12}, \quad \theta = \frac{11\pi}{12}.
   \]

2. For \( k = 1 \):
   \[
   \theta = \frac{7\pi}{12} + \pi = \frac{7\pi}{12} + \frac{12\pi}{12} = \frac{19\pi}{12},
   \]
   \[
   \theta = \frac{11\pi}{12} + \pi = \frac{11\pi}{12} + \frac{12\pi}{12} = \frac{23\pi}{12}.
   \]

### Step 4: Verify the values are within the range

Now we verify that all the values are within \( 0 \leq \theta \leq 2\pi \):

\[
0 \leq \frac{7\pi}{12} \leq 2\pi, \quad 0 \leq \frac{11\pi}{12} \leq 2\pi, \quad 0 \leq \frac{19\pi}{12} \leq 2\pi, \quad 0 \leq \frac{23\pi}{12} \leq 2\pi.
\]

All four values are within the desired range.

### Conclusion

The solutions for \( \theta \) in the interval \( 0 \leq \theta \leq 2\pi \) are:

\[
\theta = \frac{7\pi}{12}, \; \frac{11\pi}{12}, \; \frac{19\pi}{12}, \; \frac{23\pi}{12}.
\]

Would you like more details or have any questions?

### Related Questions

1. How do we solve equations involving cosine or tangent functions?
2. What are the general solutions for trigonometric equations?
3. How do the values of trigonometric functions change in different quadrants?
4. How to solve trigonometric equations using a calculator?
5. Can this problem be solved graphically?

**Tip:** Remember, for \( \sin(x) = -\frac{1}{2} \), the angles \( x \) correspond to the third and fourth quadrants of the unit circle.

Discover how to solve the equation sin(2θ) = -1/2 in the interval 0 ≤ θ ≤ 2π using trigonometric concepts. Learn the step-by-step process involving sine function values, periodicity, and angle solutions in radians. Ideal for high school students in Grades 10-12.

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