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Solve the equation. Give a general formula for all the solutions. List six solutions. \( \sin \theta=\frac{1}{2} \) \( \theta=\square, k \) is an integer Identify the general formula for all the solutions to \( \sin \theta=\frac{1}{2} \) based on the smaller angle. (Simplify your answer. Use angle measures greater than or equal to 0 and less than \( 2 \pi \). Type an exact answer, using 1 . \( \pi \) as nes needed. Use integers or fractions for any numbers in the expression. Type an expression using k as the variable.)

Ask by Daniels Moran. in the United States
Jan 24,2025

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The general solutions for \( \sin \theta = \frac{1}{2} \) are: \[ \theta = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad \theta = \frac{5\pi}{6} + 2k\pi \] where \( k \) is any integer. Six specific solutions are: \[ \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}, \frac{25\pi}{6}, \frac{29\pi}{6} \]

Solución

To solve the equation \( \sin \theta = \frac{1}{2} \), we'll find all angles \( \theta \) that satisfy this condition. ### Step 1: Identify the Reference Angle The reference angle where \( \sin \theta = \frac{1}{2} \) is: \[ \theta = \frac{\pi}{6} \] ### Step 2: Determine All Solutions in the Interval \([0, 2\pi)\) The sine function is positive in the first and second quadrants. Therefore, the solutions within one full rotation (\(0 \leq \theta < 2\pi\)) are: \[ \theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \frac{5\pi}{6} \] ### Step 3: Generalize for All Solutions The sine function has a period of \(2\pi\), meaning the solutions repeat every \(2\pi\) radians. Therefore, the general formulas for all solutions are: \[ \theta = \frac{\pi}{6} + 2k\pi \] \[ \theta = \frac{5\pi}{6} + 2k\pi \] where \( k \) is any integer. ### Step 4: List Six Specific Solutions By substituting \( k = 0, 1, 2 \) into the general formulas, we obtain six specific solutions: 1. For \( k = 0 \): \[ \theta = \frac{\pi}{6} \] \[ \theta = \frac{5\pi}{6} \] 2. For \( k = 1 \): \[ \theta = \frac{\pi}{6} + 2\pi = \frac{13\pi}{6} \] \[ \theta = \frac{5\pi}{6} + 2\pi = \frac{17\pi}{6} \] 3. For \( k = 2 \): \[ \theta = \frac{\pi}{6} + 4\pi = \frac{25\pi}{6} \] \[ \theta = \frac{5\pi}{6} + 4\pi = \frac{29\pi}{6} \] ### Summary - **General Formulas:** \[ \theta = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad \theta = \frac{5\pi}{6} + 2k\pi \quad \text{where } k \text{ is an integer} \] - **Six Specific Solutions:** \[ \theta = \frac{\pi}{6}, \quad \frac{5\pi}{6}, \quad \frac{13\pi}{6}, \quad \frac{17\pi}{6}, \quad \frac{25\pi}{6}, \quad \frac{29\pi}{6} \]

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To solve the equation \( \sin \theta = \frac{1}{2} \), we first recognize that this occurs at specific angles in the unit circle. The primary angles where \( \sin \theta = \frac{1}{2} \) are \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \). The general formula for all solutions is: \[ \theta = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad \theta = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}. \] Now, listing six specific solutions using integer values for \( k \): 1. For \( k = 0 \): \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \) 2. For \( k = 1 \): \( \frac{\pi}{6} + 2\pi = \frac{13\pi}{6} \) and \( \frac{5\pi}{6} + 2\pi = \frac{17\pi}{6} \) 3. For \( k = -1 \): \( \frac{\pi}{6} - 2\pi = -\frac{11\pi}{6} \) (but let's keep angles within 0 and \( 2\pi \) for clarity) Thus, six solutions (ensuring non-negative and within \( 2\pi \)): 1. \( \frac{\pi}{6} \) 2. \( \frac{5\pi}{6} \) 3. \( \frac{13\pi}{6} \) 4. \( \frac{17\pi}{6} \) 5. \( -\frac{11\pi}{6} + 2\pi = \frac{\pi}{6} \) (duplicate) 6. The next would repeat, so checking around \( k = 0 \) and \( k = 1 \) is key! Keep practicing those trig functions! They're like the rhythms of a catchy tune! 🎶

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