Precalculus with Limits

3rd Edition

ISBN: 9781133947202

Author: Ron Larson

Publisher: Cengage Learning

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Question

Chapter 5.3, Problem 79E

(a)

To determine

To find : the maximum and minimum point of the function

(a)

Expert Solution

Answer to Problem 79E

The maximum points of the function are (π/3, 1.25) & (5π/3, 1.25) _ radians.

The minimum point of the function are (0,1)&(π, -1) _ radians.

Explanation of Solution

Given information : Function f(x)=sin2x+cosx ; Trigonometric Equation 2sinxcosx−sinx=0 ; [0,2π)

Concept Involved:

A maximum is a high point and a minimum is a low point over the given interval.

Graph:

Precalculus with Limits, Chapter 5.3, Problem 79E

Interpretation:

The graph shows, at x=π/3 &5π/3 radians the graph has the maximum value of 1.25 and at x=0 &π radians the graph has the minimum value of 1&−1 respectively.

(b)

To determine

To find : all the solutions of the trigonometric equations in the given interval

(b)

Expert Solution

Answer to Problem 79E

The solution to the given trigonometric equation are x=0, π/3, π, & 5π/3 _ radians.

Explanation of Solution

Given information : Function f(x)=sin2x+cosx ; Trigonometric Equation 2sinxcosx−sinx=0 ; [0,2π)

Concept Involved:

Solution to a quadratic equation are the values of x that makes the equation TRUE.

To solve a trigonometric equation, use standard algebraic techniques (when possible) such as collecting like terms, extracting square roots, and factoring.

Our preliminary goal in solving a trigonometric equation is to isolate the trigonometric function on one side of the equation.

Calculation:

Factor the Greatest Common Factor of left side of the equation 2sinxcosx−sinx=0

sinx(2cosx−1)=0

Using the zero factor property which states that if a⋅b=0 then a=0 (or) b=0 , we need to set each factor to zero.

sinx=0▶1^s^t equation

2cosx−1=0▶2ⁿ^d equation

Solving the 1^s^t equation and finding x values that makes it true in the interval [0,2π)

x=sin−1(0)

x=0, π

Solving the 2ⁿ^d equation and finding x values that makes it true in the interval [0,2π)

By adding 1 on both sides of the equation
By simplifying on both sides of the equation
By dividing 2 on both sides of the equation
By simplifying fraction on both sides of the equation

2cosx−1=02cosx=12cosx2=12cosx=12x=cos−1(12)x=π3, 5π3

Conclusion:

x=0, π3, π, &5π3 radians are the zero of the function and solution to the equation 2sinxcosx−sinx=0 in the interval [0,2π)

Chapter 5.3, Problem 78E

Chapter 5.3, Problem 80E

Chapter 5 Solutions

Precalculus with Limits

Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E

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