Find all solutions of the equation in the interval [0,2). 1 2 cos 3x =
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![**Problem Statement:**
Find all solutions of the equation in the interval \([0, 2\pi)\).
\[
\cos 3x = -\frac{1}{2}
\]
**Solution Explanation:**
To solve this equation, recognize that \(\cos \theta = -\frac{1}{2}\) has solutions where the cosine function takes specific angles. The principal values where cosine equals \(-\frac{1}{2}\) are \( \theta = \frac{2\pi}{3} \) and \(\theta = \frac{4\pi}{3}\) within the interval \([0, 2\pi)\).
Given \(\cos 3x = -\frac{1}{2}\), this implies:
1. \(3x = \frac{2\pi}{3} + 2k\pi\)
2. \(3x = \frac{4\pi}{3} + 2k\pi\)
For each of these general solutions, solve for \(x\):
- From \(3x = \frac{2\pi}{3} + 2k\pi\), divide by 3:
\[
x = \frac{2\pi}{9} + \frac{2k\pi}{3}
\]
- From \(3x = \frac{4\pi}{3} + 2k\pi\), divide by 3:
\[
x = \frac{4\pi}{9} + \frac{2k\pi}{3}
\]
Find values of \(x\) that fall within the interval \([0, 2\pi)\):
1. For \(x = \frac{2\pi}{9} + \frac{2k\pi}{3}\):
- \(k = 0\): \(x = \frac{2\pi}{9}\)
- \(k = 1\): \(x = \frac{2\pi}{9} + \frac{2\pi}{3} = \frac{8\pi}{9}\)
- \(k = 2\): \(x = \frac{2\pi}{9} + \frac{4\pi}{3} = \frac{14\pi}{9}\)
- \(k = 3\): \(x = \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbcea6d79-c0c2-49cd-aa31-c3eca199518f%2Fb4f542cf-abfc-44f8-b631-e490c53d7c6f%2Feydk0yc_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find all solutions of the equation in the interval \([0, 2\pi)\).
\[
\cos 3x = -\frac{1}{2}
\]
**Solution Explanation:**
To solve this equation, recognize that \(\cos \theta = -\frac{1}{2}\) has solutions where the cosine function takes specific angles. The principal values where cosine equals \(-\frac{1}{2}\) are \( \theta = \frac{2\pi}{3} \) and \(\theta = \frac{4\pi}{3}\) within the interval \([0, 2\pi)\).
Given \(\cos 3x = -\frac{1}{2}\), this implies:
1. \(3x = \frac{2\pi}{3} + 2k\pi\)
2. \(3x = \frac{4\pi}{3} + 2k\pi\)
For each of these general solutions, solve for \(x\):
- From \(3x = \frac{2\pi}{3} + 2k\pi\), divide by 3:
\[
x = \frac{2\pi}{9} + \frac{2k\pi}{3}
\]
- From \(3x = \frac{4\pi}{3} + 2k\pi\), divide by 3:
\[
x = \frac{4\pi}{9} + \frac{2k\pi}{3}
\]
Find values of \(x\) that fall within the interval \([0, 2\pi)\):
1. For \(x = \frac{2\pi}{9} + \frac{2k\pi}{3}\):
- \(k = 0\): \(x = \frac{2\pi}{9}\)
- \(k = 1\): \(x = \frac{2\pi}{9} + \frac{2\pi}{3} = \frac{8\pi}{9}\)
- \(k = 2\): \(x = \frac{2\pi}{9} + \frac{4\pi}{3} = \frac{14\pi}{9}\)
- \(k = 3\): \(x = \
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