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Solve the equation. Give a general formula for all the solutions. List six solutions. \( \sin \theta=-\frac{\sqrt{3}}{2} \) Identify the general formula for all the solutions to \( \sin \theta=-\frac{\sqrt{3}}{2} \) based on the smaller angle. \( \theta=\square, \mathrm{k} \) is an integer (Simplify your answer. Use angle measures greater than or equal to 0 and less than \( 2 \pi \). Type an exact answer, using \( \pi \) as needed. Use integers or fractions for any numbers in the expression. Type an expression using k as the variable.)

Ask by Conner Lyons. in the United States
Jan 24,2025

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**General Solution:** \[ \theta = \frac{4\pi}{3} + 2\pi k \quad \text{and} \quad \theta = \frac{5\pi}{3} + 2\pi k \quad \text{where } k \text{ is an integer} \] **Six Specific Solutions:** \[ \frac{4\pi}{3}, \quad \frac{5\pi}{3}, \quad \frac{10\pi}{3}, \quad \frac{11\pi}{3}, \quad \frac{16\pi}{3}, \quad \frac{17\pi}{3} \]

Solución

To solve the equation \( \sin \theta = -\frac{\sqrt{3}}{2} \), we'll follow these steps: ### 1. **Determine the Reference Angle** The reference angle \( \alpha \) where \( \sin \alpha = \frac{\sqrt{3}}{2} \) is: \[ \alpha = \frac{\pi}{3} \] ### 2. **Find the Quadrants Where Sine is Negative** Sine is negative in the **third** and **fourth** quadrants. Therefore, the solutions within one full rotation (\( 0 \) to \( 2\pi \)) are: \[ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \quad \text{(Third Quadrant)} \] \[ \theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \quad \text{(Fourth Quadrant)} \] ### 3. **Write the General Solution** Since the sine function has a period of \( 2\pi \), the general solutions are: \[ \theta = \frac{4\pi}{3} + 2\pi k \quad \text{or} \quad \theta = \frac{5\pi}{3} + 2\pi k \quad \text{where } k \text{ is any integer} \] ### 4. **List Six Specific Solutions** To list six solutions, we can choose different integer values for \( k \): 1. For \( k = 0 \): \[ \theta = \frac{4\pi}{3}, \quad \frac{5\pi}{3} \] 2. For \( k = 1 \): \[ \theta = \frac{4\pi}{3} + 2\pi = \frac{10\pi}{3}, \quad \frac{5\pi}{3} + 2\pi = \frac{11\pi}{3} \] 3. For \( k = 2 \): \[ \theta = \frac{4\pi}{3} + 4\pi = \frac{16\pi}{3}, \quad \frac{5\pi}{3} + 4\pi = \frac{17\pi}{3} \] ### **Final Answer** **General Formula:** \[ \theta = \frac{4\pi}{3} + 2\pi k \quad \text{and} \quad \theta = \frac{5\pi}{3} + 2\pi k \quad \text{where } k \text{ is an integer} \] **Six Specific Solutions:** \[ \frac{4\pi}{3}, \quad \frac{5\pi}{3}, \quad \frac{10\pi}{3}, \quad \frac{11\pi}{3}, \quad \frac{16\pi}{3}, \quad \frac{17\pi}{3} \]

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To solve the equation \( \sin \theta = -\frac{\sqrt{3}}{2} \), we first identify the angles where the sine function takes on that value. The angle \( \theta \) that has a sine of \( -\frac{\sqrt{3}}{2} \) corresponds to the reference angle of \( \frac{\pi}{3} \) in the fourth and third quadrants. Therefore, the general formulas for the solutions are: \[ \theta = \frac{4\pi}{3} + 2k\pi \quad \text{and} \quad \theta = \frac{5\pi}{3} + 2k\pi \] where \( k \) is any integer. Now, let's list six specific solutions by substituting different integer values for \( k \): 1. For \( k = 0 \): \( \theta = \frac{4\pi}{3}, \frac{5\pi}{3} \) 2. For \( k = 1 \): \( \theta = \frac{4\pi}{3} + 2\pi = \frac{10\pi}{3}, \frac{5\pi}{3} + 2\pi = \frac{11\pi}{3} \) 3. For \( k = -1 \): \( \theta = \frac{4\pi}{3} - 2\pi = -\frac{2\pi}{3}, \frac{5\pi}{3} - 2\pi = -\frac{1\pi}{3} \) So, the six solutions we have identified are: \( \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{10\pi}{3}, \frac{11\pi}{3}, -\frac{2\pi}{3}, -\frac{1\pi}{3} \)

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