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Find all solutions of the equation 2 sin x V3 = 0 where 0 < x < 2T - The answers are

Find all solutions of the equation 2 sin x V3 = 0 where 0 < x < 2T - The answers are

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.5: Product-to-sum And Sum-to-product Formulas
Problem 2E
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Question
**Problem Statement** Find all solutions of the equation \(2 \sin x - \sqrt{3} = 0\) where \(0 < x < 2\pi\). **Solution** To solve the equation \(2 \sin x - \sqrt{3} = 0\), we need to isolate \(\sin x\): 1. \(2 \sin x = \sqrt{3}\) 2. \(\sin x = \frac{\sqrt{3}}{2}\) The solutions for \(\sin x = \frac{\sqrt{3}}{2}\) within the interval \(0 < x < 2\pi\) are: - \(x = \frac{\pi}{3}\) - \(x = \frac{2\pi}{3}\) **Final Answer** The answers are \(\frac{\pi}{3}\) and \(\frac{2\pi}{3}\).
Transcribed Image Text:**Problem Statement** Find all solutions of the equation \(2 \sin x - \sqrt{3} = 0\) where \(0 < x < 2\pi\). **Solution** To solve the equation \(2 \sin x - \sqrt{3} = 0\), we need to isolate \(\sin x\): 1. \(2 \sin x = \sqrt{3}\) 2. \(\sin x = \frac{\sqrt{3}}{2}\) The solutions for \(\sin x = \frac{\sqrt{3}}{2}\) within the interval \(0 < x < 2\pi\) are: - \(x = \frac{\pi}{3}\) - \(x = \frac{2\pi}{3}\) **Final Answer** The answers are \(\frac{\pi}{3}\) and \(\frac{2\pi}{3}\).
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