   Chapter 8.2, Problem 64E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
1 views

# Solving a Trigonometric Equation In Exercises 61-70, solve the equation for θ . Assume 0 ≤ θ ≤ 2 π . For some of the equations, you should use the trigonometric identities listed in this section. Use a spreadsheet or a graphing utility to verify your results. See Example 7. 2 cos 2 θ − cos θ = 1

To determine

The values of θ for the trigonometric equation 2cos2θcosθ=1 and verify the result by graphing utility tool.

Explanation

Given Information:

The provided trigonometric equation is, 2cos2θcosθ=1.

Consider the trigonometric equation, 2cos2θcosθ=1

Simplify the provided trigonometry equation,

2cos2θcosθ=12cos2θcosθ1=02cos2θ2cosθ+cosθ1=0(2cosθ+1)(cosθ1)=0

Further solve,

2cosθ+1=02cosθ=1cosθ=12

And,

cosθ1=0cosθ=1

Since, cosine is positive in first and forth quadrants, while cosine is negative in second and third quadrants,

Thus, there are four solutions for trigonometric equation s 2cos2θcosθ=1 for 0θ2π, which can be mathematically determined as:

First value:

cosθ=1cosθ=cos0θ=0,

Second value:

cosθ=1cosθ=cos2πθ=2π=6.283185301

Third value:

cosθ=12cosθ=cos2π3θ=2π3=2.094395102

Fourth value:

cosθ=12cosθ=cos4π3θ=4π3=4.18879020

Therefore, the values of θ for the trigonometric equation 2cos2θcosθ=1 are θ=0, θ=2π3, θ=4π3 and θ=2π

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