Solve the following inequality. Express the exact answer in interval notation, restricting your attention to - 2m ≤x≤ 2개. tan2(x) > 1 U 아래 - 피아피아 이제 4 O 아주 4 ○ [-2x2x] 아주 - U 아주아주 U 3개 4 U -1 아주 5개 7개" (-을-위플)를 뛰며)(플 퓌 -플) 아플 - 피아 O [-27 - 피아뜸 -x) U (-7-24] 아픔) (0] 2' U x) (4] [ 2] U
Solve the following inequality. Express the exact answer in interval notation, restricting your attention to - 2m ≤x≤ 2개. tan2(x) > 1 U 아래 - 피아피아 이제 4 O 아주 4 ○ [-2x2x] 아주 - U 아주아주 U 3개 4 U -1 아주 5개 7개" (-을-위플)를 뛰며)(플 퓌 -플) 아플 - 피아 O [-27 - 피아뜸 -x) U (-7-24] 아픔) (0] 2' U x) (4] [ 2] U
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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![### Solving the Inequality for Tangent Squared
To solve the inequality \( \tan^2(x) \geq 1 \) within the interval \(-2\pi \leq x \leq 2\pi \), we must express the exact answer using interval notation.
Given the options, we need to identify where the tangent function satisfies the inequality:
\[ \tan^2(x) \geq 1 \]
This is equivalent to:
\[ |\tan(x)| \geq 1 \]
Which in turn implies:
\[ \tan(x) \leq -1 \quad \text{or} \quad \tan(x) \geq 1 \]
The tangent function has a period of \(\pi\). The principal solutions where \(\tan(x) = 1\) are at \(x = \frac{\pi}{4} + n\pi\), and for \(\tan(x) = -1\) at \(x = -\frac{\pi}{4} + n\pi\).
Considering the given interval \(-2\pi \leq x \leq 2\pi\), we look at all intervals that match \(|\tan(x)| \geq 1\):
The correct interval representation is:
\[ [-2\pi, -\frac{7\pi}{4}) \cup [-\frac{5\pi}{4}, -\frac{3\pi}{4}) \cup [-\frac{\pi}{4}, \frac{\pi}{4}] \cup (\frac{3\pi}{4}, \frac{5\pi}{4}] \cup (\frac{7\pi}{4}, 2\pi] \]
This matches one of the choices given:
\[ \begin{aligned}
& -2\pi \leq x < -\frac{7\pi}{4}, \\
& -\frac{5\pi}{4} \leq x < -\frac{3\pi}{4}, \\
& -\frac{\pi}{4} \leq x \leq \frac{\pi}{4},\\
& \frac{3\pi}{4} < x \leq \frac{5\pi}{4},\\
& \frac{7\pi}{4} < x \leq 2\pi.
\end{aligned} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13d72ac4-a1a6-4f9c-b702-17497af57bfa%2F216cb240-b6bd-4efa-91f8-6973791ce1a3%2F81oyd2g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving the Inequality for Tangent Squared
To solve the inequality \( \tan^2(x) \geq 1 \) within the interval \(-2\pi \leq x \leq 2\pi \), we must express the exact answer using interval notation.
Given the options, we need to identify where the tangent function satisfies the inequality:
\[ \tan^2(x) \geq 1 \]
This is equivalent to:
\[ |\tan(x)| \geq 1 \]
Which in turn implies:
\[ \tan(x) \leq -1 \quad \text{or} \quad \tan(x) \geq 1 \]
The tangent function has a period of \(\pi\). The principal solutions where \(\tan(x) = 1\) are at \(x = \frac{\pi}{4} + n\pi\), and for \(\tan(x) = -1\) at \(x = -\frac{\pi}{4} + n\pi\).
Considering the given interval \(-2\pi \leq x \leq 2\pi\), we look at all intervals that match \(|\tan(x)| \geq 1\):
The correct interval representation is:
\[ [-2\pi, -\frac{7\pi}{4}) \cup [-\frac{5\pi}{4}, -\frac{3\pi}{4}) \cup [-\frac{\pi}{4}, \frac{\pi}{4}] \cup (\frac{3\pi}{4}, \frac{5\pi}{4}] \cup (\frac{7\pi}{4}, 2\pi] \]
This matches one of the choices given:
\[ \begin{aligned}
& -2\pi \leq x < -\frac{7\pi}{4}, \\
& -\frac{5\pi}{4} \leq x < -\frac{3\pi}{4}, \\
& -\frac{\pi}{4} \leq x \leq \frac{\pi}{4},\\
& \frac{3\pi}{4} < x \leq \frac{5\pi}{4},\\
& \frac{7\pi}{4} < x \leq 2\pi.
\end{aligned} \]
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