Find all solutions in radians using exact values only: sin² (8x) = 1
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°.
Related questions
Question
![**Problem 9:**
Find all solutions in radians using exact values only: \( \sin^2(8x) = 1 \)
**Explanation:**
To solve this equation, notice that \( \sin^2(8x) = 1 \) implies that \( \sin(8x) = \pm1 \). This occurs when \( 8x \) is an odd multiple of \( \frac{\pi}{2} \). Specifically,
\[
8x = \frac{\pi}{2} + k\pi,
\]
where \( k \) is an integer. Solving for \( x \), we have:
\[
x = \frac{\pi}{16} + \frac{k\pi}{8}
\]
Hence, the solutions are in the form:
\[
x = \frac{\pi}{16} + \frac{k\pi}{8}, \, k \in \mathbb{Z}.
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Transcribed Image Text:**Problem 9:**
Find all solutions in radians using exact values only: \( \sin^2(8x) = 1 \)
**Explanation:**
To solve this equation, notice that \( \sin^2(8x) = 1 \) implies that \( \sin(8x) = \pm1 \). This occurs when \( 8x \) is an odd multiple of \( \frac{\pi}{2} \). Specifically,
\[
8x = \frac{\pi}{2} + k\pi,
\]
where \( k \) is an integer. Solving for \( x \), we have:
\[
x = \frac{\pi}{16} + \frac{k\pi}{8}
\]
Hence, the solutions are in the form:
\[
x = \frac{\pi}{16} + \frac{k\pi}{8}, \, k \in \mathbb{Z}.
\]
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