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

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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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