/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 99 Determine whether the statement ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 99

Determine whether the statement is true or false. Justify your answer. The equation \(2 \sin 4 t-1=0\) has four times the number of solutions in the interval \([0,2 \pi)\) as the equation \(2 \sin t-1=0\).

Short Answer

Expert verified
The statement is true. The equation \(2 \sin 4 t - 1 = 0\) indeed has four times the number of solutions in the interval \([0,2 \pi)\) as the equation \(2 \sin t - 1 = 0\).

Step by step solution

01

Calculate number of solutions for \(2 \sin t - 1 = 0\)

First, solve the equation \(2 \sin t - 1 = 0\) for \(t\). Doing so gives \(t = \sin^{-1}\left(\frac{1}{2}\right) + 2 \pi k\) and \(t = \pi - \sin^{-1}\left(\frac{1}{2}\right) + 2 \pi k\) where \(k\) is integer. In the interval \([0,2 \pi)\), there are two solutions which are \(t = \frac{\pi}{6}, \frac{5\pi}{6}\).
02

Calculate number of solutions for \(2 \sin 4 t - 1 = 0\)

Next, solve the equation \(2 \sin 4t - 1 = 0\) for \(t\). This gives \(t = \frac{1}{4}\sin^{-1}\left(\frac{1}{2}\right) + \frac{\pi k}{2}\) and \(t = \frac{1}{4}(\pi - \sin^{-1}\left(\frac{1}{2}\right)) + \frac{\pi k}{2}\) where \(k\) is integer. In the interval \([0,2 \pi)\), there are indeed eight solutions which are \(t = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}\).
03

Compare the numbers of solutions

Finally, compare the results of Steps 1 and 2. The equation \(2 \sin 4 t - 1 = 0\) has eight solutions and the equation \(2 \sin t - 1 = 0\) has two solutions in the interval \([0,2 \pi)\). Therefore the equation \(2 \sin 4 t - 1 = 0\) indeed has four times the number of solutions in the interval \([0,2 \pi)\) as the equation \(2 \sin t - 1 = 0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find all solutions of the equation in the interval \([0,2 \pi)\). $$\cos (x+\pi)-\cos x-1=0$$

Use a graphing utility to approximate the solutions of the equation in the interval \([0,2 \pi)\). $$\cos \left(x-\frac{\pi}{2}\right)-\sin ^{2} x=0$$

Verify the identity. $$\frac{\sin x \pm \sin y}{\cos x+\cos y}=\tan \frac{x \pm y}{2}$$

Find all solutions of the equation in the interval \([0,2 \pi) .\) Use a graphing utility to graph the equation and verify the solutions. $$\cos \frac{x}{2}-\sin x=0$$

Use the power-reducing formulas to rewrite the expression in terms of the first power of the cosine. $$\sin ^{2} 2 x \cos ^{2} 2 x$$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.