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Enter all solutions, in degrees, that lie in the interval [0°, 360°). Do not enter the degree sign. If there are more than one angle, separate them by a comma. 2 sin y+ 3 siny-2 = 0

Enter all solutions, in degrees, that lie in the interval [0°, 360°). Do not enter the degree sign. If there are more than one angle, separate them by a comma. 2 sin y+ 3 siny-2 = 0

Trigonometry (MindTap Course List)
Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter2: Right Triangle Trigonometry
Section: Chapter Questions
Problem 5GP
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### Solving Trigonometric Equations: Example Problem **Problem Statement:** Solve the equation given. \[ 2 \sin^2 \gamma + 3 \sin \gamma - 2 = 0 \] Enter all solutions, in degrees, that lie in the interval \([0^\circ, 360^\circ)\). Do not enter the degree sign. If there are more than one angle, separate them by a comma. \[ \gamma = \] **Instructions for Students:** 1. Identify the type of trigonometric equation. 2. Use algebraic techniques to solve for the variable within the specified interval. 3. Ensure that all solutions are within the given range and in degrees. **Example Solution Process:** 1. Start by rewriting the equation: \[ 2 \sin^2 \gamma + 3 \sin \gamma - 2 = 0 \] 2. Factor the quadratic equation in \(\sin \gamma\): \[ (2 \sin \gamma - 1)(\sin \gamma + 2) = 0 \] 3. Solve for \(\sin \gamma\) for each factor: \[ 2 \sin \gamma - 1 = 0 \quad \Rightarrow \quad \sin \gamma = \frac{1}{2} \] \[ \sin \gamma + 2 = 0 \quad \Rightarrow \quad \sin \gamma = -2 \] (Note: \(\sin \gamma = -2\) is not possible as the sine of an angle cannot be less than -1 or greater than 1.) 4. Solve for \(\gamma\) when \(\sin \gamma = \frac{1}{2}\): - \(\gamma = 30^\circ\) - \(\gamma = 150^\circ\) (because \(\sin \gamma = \frac{1}{2}\) in both the first and second quadrants) 5. Therefore, the solutions within the interval \([0^\circ, 360^\circ)\) are: \[ \gamma = 30, 150 \] 6. Enter these solutions separated by commas (without the degree sign). \[ \gamma = 30, 150 \] By following these steps, you will arrive at the correct solutions.
Transcribed Image Text:### Solving Trigonometric Equations: Example Problem **Problem Statement:** Solve the equation given. \[ 2 \sin^2 \gamma + 3 \sin \gamma - 2 = 0 \] Enter all solutions, in degrees, that lie in the interval \([0^\circ, 360^\circ)\). Do not enter the degree sign. If there are more than one angle, separate them by a comma. \[ \gamma = \] **Instructions for Students:** 1. Identify the type of trigonometric equation. 2. Use algebraic techniques to solve for the variable within the specified interval. 3. Ensure that all solutions are within the given range and in degrees. **Example Solution Process:** 1. Start by rewriting the equation: \[ 2 \sin^2 \gamma + 3 \sin \gamma - 2 = 0 \] 2. Factor the quadratic equation in \(\sin \gamma\): \[ (2 \sin \gamma - 1)(\sin \gamma + 2) = 0 \] 3. Solve for \(\sin \gamma\) for each factor: \[ 2 \sin \gamma - 1 = 0 \quad \Rightarrow \quad \sin \gamma = \frac{1}{2} \] \[ \sin \gamma + 2 = 0 \quad \Rightarrow \quad \sin \gamma = -2 \] (Note: \(\sin \gamma = -2\) is not possible as the sine of an angle cannot be less than -1 or greater than 1.) 4. Solve for \(\gamma\) when \(\sin \gamma = \frac{1}{2}\): - \(\gamma = 30^\circ\) - \(\gamma = 150^\circ\) (because \(\sin \gamma = \frac{1}{2}\) in both the first and second quadrants) 5. Therefore, the solutions within the interval \([0^\circ, 360^\circ)\) are: \[ \gamma = 30, 150 \] 6. Enter these solutions separated by commas (without the degree sign). \[ \gamma = 30, 150 \] By following these steps, you will arrive at the correct solutions.
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