Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Solving \( \tan^2 x \sin x = \sin x \) on the Interval \([0, 2\pi]\)
To solve the equation \(\tan^2 x \sin x = \sin x\) on the interval \([0, 2\pi]\), we need to determine the values of \(x\) where this equation holds true.
Given the equation:
\[ \tan^2 x \sin x = \sin x \]
First, we can factor out the \(\sin x\):
\[ \sin x (\tan^2 x - 1) = 0 \]
This equation is satisfied when:
1. \(\sin x = 0\)
2. \(\tan^2 x - 1 = 0\)
Let's solve each case separately:
#### Case 1: \(\sin x = 0\)
\(\sin x = 0\) when \(x = 0, \pi, 2\pi\) in the interval \([0, 2\pi]\).
#### Case 2: \(\tan^2 x - 1 = 0\)
\(\tan^2 x = 1\) implies \(\tan x = \pm 1\).
\(\tan x = 1\) at \(x = \frac{\pi}{4}, \frac{5\pi}{4}\)
\(\tan x = -1\) at \(x = \frac{3\pi}{4}, \frac{7\pi}{4}\)
So the solutions in this interval are:
\[ x = 0, \frac{\pi}{4}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{7\pi}{4}, 2\pi \]
### Options Assessment
Presented options are:
1. \( \frac{\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4} \)
2. \( \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}, 0, \pi, 2\pi \)
3. \( 0, \pi, 2\pi \)
4. \( \frac{\pi](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f02248d-9b04-4bf9-ba6b-18c74681389d%2F5fa045b5-7ab6-473c-a155-bf028b07df6a%2Fjhw1rw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving \( \tan^2 x \sin x = \sin x \) on the Interval \([0, 2\pi]\)
To solve the equation \(\tan^2 x \sin x = \sin x\) on the interval \([0, 2\pi]\), we need to determine the values of \(x\) where this equation holds true.
Given the equation:
\[ \tan^2 x \sin x = \sin x \]
First, we can factor out the \(\sin x\):
\[ \sin x (\tan^2 x - 1) = 0 \]
This equation is satisfied when:
1. \(\sin x = 0\)
2. \(\tan^2 x - 1 = 0\)
Let's solve each case separately:
#### Case 1: \(\sin x = 0\)
\(\sin x = 0\) when \(x = 0, \pi, 2\pi\) in the interval \([0, 2\pi]\).
#### Case 2: \(\tan^2 x - 1 = 0\)
\(\tan^2 x = 1\) implies \(\tan x = \pm 1\).
\(\tan x = 1\) at \(x = \frac{\pi}{4}, \frac{5\pi}{4}\)
\(\tan x = -1\) at \(x = \frac{3\pi}{4}, \frac{7\pi}{4}\)
So the solutions in this interval are:
\[ x = 0, \frac{\pi}{4}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{7\pi}{4}, 2\pi \]
### Options Assessment
Presented options are:
1. \( \frac{\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4} \)
2. \( \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}, 0, \pi, 2\pi \)
3. \( 0, \pi, 2\pi \)
4. \( \frac{\pi
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