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Question 2 > Find all solutions of the equation sec² X 2 = 0. π The answer is A + Bkπ where k is any integer and 0 < A < 2 A B = = 2 2

Question 2 > Find all solutions of the equation sec² X 2 = 0. π The answer is A + Bkπ where k is any integer and 0 < A < 2 A B = = 2 2

Calculus: Early Transcendentals
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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### Question 2 Find all solutions of the equation sec^2(x) - 2 = 0. The answer is \( A + Bk\pi \) where \( k \) is any integer and \( 0 < A < \frac{\pi}{2} \), \[ A = \quad , \quad B = \] This problem requires you to find the values of \( A \) and \( B \) that satisfy the given equation. The solution should have a specific format where \( A \) is a constant within the specified range, and \( B \) involves \( k \), where \( k \) is any integer. ---
Transcribed Image Text:### Question 2 Find all solutions of the equation sec^2(x) - 2 = 0. The answer is \( A + Bk\pi \) where \( k \) is any integer and \( 0 < A < \frac{\pi}{2} \), \[ A = \quad , \quad B = \] This problem requires you to find the values of \( A \) and \( B \) that satisfy the given equation. The solution should have a specific format where \( A \) is a constant within the specified range, and \( B \) involves \( k \), where \( k \) is any integer. ---
### Trigonometric Identities and Calculations #### Problem Statement: If \( \sin x = -\frac{3}{5} \), \( x \) in quadrant III, then: - \( \sin 2x = \underline{\hspace{50pt}} ; \) - \( \cos 2x = \underline{\hspace{50pt}} ; \) - \( \tan 2x = \underline{\hspace{50pt}} .\) #### Explanation: This problem involves applying trigonometric identities to find the values of \( \sin 2x \), \( \cos 2x \), and \( \tan 2x \), given that \( \sin x = -\frac{3}{5} \) and \( x \) is in the third quadrant. In the third quadrant, sine is negative and cosine is also negative. **1. Find \( \cos x \):** Using the Pythagorean identity: \[ \sin^2 x + \cos^2 x = 1 \] Given: \[ \sin x = -\frac{3}{5} \] We square it: \[ \left(-\frac{3}{5}\right)^2 + \cos^2 x = 1 \] \[ \frac{9}{25} + \cos^2 x = 1 \] \[ \cos^2 x = 1 - \frac{9}{25} \] \[ \cos^2 x = \frac{25}{25} - \frac{9}{25} \] \[ \cos^2 x = \frac{16}{25} \] Since \( x \) is in the third quadrant, \( \cos x \) is negative: \[ \cos x = -\frac{4}{5} \] **2. Calculate \( \sin 2x \):** Using the double-angle identity for sine: \[ \sin 2x = 2 \sin x \cos x \] \[ \sin 2x = 2 \left(-\frac{3}{5}\right) \left(-\frac{4}{5}\right) \] \[ \sin 2x = 2 \cdot \frac{12}{25} \] \[ \sin 2x = \frac{24}{25} \] **3. Calculate \( \cos 2x \):
Transcribed Image Text:### Trigonometric Identities and Calculations #### Problem Statement: If \( \sin x = -\frac{3}{5} \), \( x \) in quadrant III, then: - \( \sin 2x = \underline{\hspace{50pt}} ; \) - \( \cos 2x = \underline{\hspace{50pt}} ; \) - \( \tan 2x = \underline{\hspace{50pt}} .\) #### Explanation: This problem involves applying trigonometric identities to find the values of \( \sin 2x \), \( \cos 2x \), and \( \tan 2x \), given that \( \sin x = -\frac{3}{5} \) and \( x \) is in the third quadrant. In the third quadrant, sine is negative and cosine is also negative. **1. Find \( \cos x \):** Using the Pythagorean identity: \[ \sin^2 x + \cos^2 x = 1 \] Given: \[ \sin x = -\frac{3}{5} \] We square it: \[ \left(-\frac{3}{5}\right)^2 + \cos^2 x = 1 \] \[ \frac{9}{25} + \cos^2 x = 1 \] \[ \cos^2 x = 1 - \frac{9}{25} \] \[ \cos^2 x = \frac{25}{25} - \frac{9}{25} \] \[ \cos^2 x = \frac{16}{25} \] Since \( x \) is in the third quadrant, \( \cos x \) is negative: \[ \cos x = -\frac{4}{5} \] **2. Calculate \( \sin 2x \):** Using the double-angle identity for sine: \[ \sin 2x = 2 \sin x \cos x \] \[ \sin 2x = 2 \left(-\frac{3}{5}\right) \left(-\frac{4}{5}\right) \] \[ \sin 2x = 2 \cdot \frac{12}{25} \] \[ \sin 2x = \frac{24}{25} \] **3. Calculate \( \cos 2x \):
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