/*! 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 5 Solve the equation $$ \cos x... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 5

Solve the equation $$ \cos x+\cos 2 x-\cos 3 x=1 $$

Short Answer

Expert verified
Answer: The values of x can be determined by solving the equation with the help of trigonometric identities and substitutions. Since the resulting cubic equation has no elementary closed-form solutions, we can continue either by solving it numerically or by identifying specific angles that would satisfy the equation. The final step is to revert back from the obtained values of y (if any) to find the desired x values.

Step by step solution

01

Write down the given equation

The given equation is: $$ \cos x + \cos 2x - \cos 3x = 1 $$
02

Substitute the Double Angle Formula for \(\cos 2x\)

We'll use the identity \(\cos 2x = 2 \cos^2 x - 1\) to replace \(\cos 2x\) in the given equation: $$ \cos x + (2 \cos^2 x - 1) - \cos 3x = 1 $$
03

Rearrange the equation to group cosine terms

Move all cosine terms to one side and all constants to the other side: $$ 2 \cos^2 x - \cos 3x + \cos x - 2 = 0 $$
04

Use the angle sum formula for \(\cos 3x\)

\(\cos 3x\) can be rewritten as \(\cos{(2x + x)}\). Using the angle sum formula, this term can be expressed as: $$ \cos 3x = \cos (2x + x) = \cos 2x \cos x - \sin 2x \sin x $$ Now substitute \(\cos 2x = 2 \cos^2 x - 1\) and \(\sin 2x = 2 \sin x \cos x\): $$ \cos 3x = (2 \cos^2 x - 1) \cos x - 2 \sin x \cos x \sin x $$
05

Substitute \(\cos 3x\) back into the equation

Replace \(\cos 3x\) with the expanded expression we found in the previous step: $$ 2 \cos^2 x - ((2 \cos^2 x - 1) \cos x - 2 \sin x \cos x \sin x) + \cos x - 2 = 0 $$
06

Introduce a substitution

Define \(y = \cos x\) and \(z = \sin x\), then the equation becomes: $$ 2y^2 - ((2y^2 - 1) y - 2z y z) + y - 2 = 0 $$
07

Solve the simplified equation for y

Work through the algebra to solve for y: $$ 2y^2 - 2y^3 + y - 2 = 0 $$ This equation has no elementary closed-form solutions for \(y\). So we can continue by either solving the cubic equation numerically or attempting to identify specific angles \(x\) that would satisfy the equation. The latter technique would involve searching for any special angles of \(x\) where \(y = \cos x\) could satisfy this equation.
08

Revert back to x

Once we find the possible value(s) of \(y\), we can revert back to \(x\) by solving the equation: $$ \cos x = y $$ The solutions to this equation will be our desired values of \(x\) that satisfy the original equation.

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

A triangle \(A_{1} A_{2} A_{3}\) and a point \(P_{0}\) are given in the plane. We define \(A_{s}=A_{s-3}\) for all \(s \geq 4 .\) We construct a sequence of points \(P_{0}, P_{1}, P_{2}, \ldots\) such that \(P_{k+1}\) is the image of \(P_{k}\) under the rotation with center \(A_{k+1}\) through the angle \(120^{\circ}\) clockwise \((k=0,1,2, \ldots) .\) Prove that if \(P_{1986}=P_{0}\), then the triangle \(A_{1} A_{2} A_{3}\) is equilateral.

Let \(n\) be an odd positive integer and \(\varepsilon_{0}, \varepsilon_{1}, \ldots, \varepsilon_{n-1}\) the complex roots of unity of order \(n .\) Prove that $$ \prod_{k=0}^{n-1}\left(a+b \varepsilon_{k}^{2}\right)=a^{n}+b^{n} $$ for all complex numbers a and \(b\).

Let \(A, B, C\) be three consecutive vertices of a regular \(n\) -gon and consider the point \(M\) on the circumcircle such that points \(B\) and \(M\) lie on opposite sides of the line \(A C\). Prove that \(M A+M C=2 M B \cos \frac{\pi}{n}\)

Let \(n \geq 3\) be an integer and \(z=\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n} .\) Consider the sets $$ A=\left\\{1, z, z^{2}, \ldots, z^{n-1}\right\\} $$ and $$ B=\left\\{1,1+z, 1+z+z^{2}, \ldots, 1+z+\ldots+z^{n-1}\right\\} . $$ Determine \(A \cap B\).

Consider a complex number \(z, z \neq 0\), and the real sequence $$ a_{n}=\left|z^{n}+\frac{1}{z^{n}}\right|, n \geq 1 . $$ (a) Show that if \(a_{1}>2\), then $$ a_{n+1}<\frac{a_{n}+a_{n+2}}{2}, \text { for all } n \in \mathbb{N}^{*} \text { . } $$ (b) Prove that if there exists \(k \in \mathbb{N}^{*}\) such that \(a_{k} \leq 2\), then \(a_{1} \leq 2\).
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Discrete 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.