/*! 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 \(2\sin t -1=0\), i.e., it has 8 solutions compared to 2 solutions of the other equation.

Step by step solution

01

Solve the first equation

Given the equation \(2 \sin t - 1 = 0\), we will first solve for \(t\) by isolating the sine function, and then applying the inverse sine function to both sides. Solving in this way provides us with \(t = \sin^{-1}(1/2)\). Using a known trigonometric result, we find \(t = \pi/6\) or \(t = 5\pi/6\). Therefore, in the interval \([0, 2\pi]\), this equation has two roots.
02

Solve the second equation

The equation is \(2 \sin 4t - 1 = 0\). Solving for \(t\) in the same way as in Step 1, we end up with \(\sin 4t = 1/2\), which means \(4t = \sin^{-1}(1/2)\). This gives values \(4t = \pi/6, 5\pi/6\). However, because we have compressed the wave four times by multiplying \(t\) by \(4\), each solution to the first equation will correspond to four solutions to this equation in the interval \([0, 2\pi]\). Therefore, this equation has 4 * 2 = 8 solutions in this interval.
03

Compare the number of solutions

We found that the first equation has 2 solutions and the second one has 8 solutions. Therefore, the statement that the second equation has four times the number of solutions as the first equation in the interval \([0, 2\pi])\) is true.

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 the exact value of each expression. (a) \(\cos \left(\frac{\pi}{4}+\frac{\pi}{3}\right)\) (b) \(\cos \frac{\pi}{4}+\cos \frac{\pi}{3}\)

Find the exact values of the sine, cosine, and tangent of the angle. $$165^{\circ}=135^{\circ}+30^{\circ}$$

Consider the function given by \(f(x)=3 \sin (0.6 x-2)\). (a) Approximate the zero of the function in the interval [0,6] (b) A quadratic approximation agreeing with \(f\) at \(x=5\) is \(g(x)=-0.45 x^{2}+5.52 x-13.70 .\) Use a graphing utility to graph \(f\) and \(g\) in the same viewing window. Describe the result. (c) Use the Quadratic Formula to find the zeros of \(g\). Compare the zero in the interval [0,6] with the result of part (a).

The monthly sales \(S\) (in thousands of units) of a seasonal product are approximated by $$S=74.50+43.75 \sin \frac{\pi t}{6}$$ where \(t\) is the time (in months), with \(t=1\) corresponding to January. Determine the months in which sales exceed 100,000 units.

Use the Quadratic Formula to solve the equation in the interval \([0,2 \pi)\). Then use a graphing utility to approximate the angle \(x\). $$\tan ^{2} x+3 \tan x+1=0$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

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.