EBK PRECALCULUS W/LIMITS

3rd Edition

ISBN: 9781285607160

Author: Larson

Publisher: CENGAGE CO

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Vertices Of An Ellipse

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Question

Chapter 5.3, Problem 82E

(a)

To determine

To find : the maximum and minimum point of the function

(a)

Expert Solution

Answer to Problem 82E

The maximum points of the function are (π/6, 1.5) & (5π/6, 1.5) _ radians.

The minimum point of the function are (π/2, 1) & (3π/2, -3) _ radians.

Explanation of Solution

Given information : Function f(x)=2sinx+cos2x ; Trigonometric Equation 2cosx−4sinxcosx=0 ; [0,2π)

Concept Involved:

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

Graph:

EBK PRECALCULUS W/LIMITS, Chapter 5.3, Problem 82E

Interpretation:

The graph of the function f(x)=2sinx+cos2x shows, at x=π6, 5π6 radians the graph has the maximum value of 1.5 and at x=π2, 3π2 radians the graph has the minimum value of 1 & −3 respectively.

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 82E

The solution to the given trigonometric equation are x= π6, π2, 5π6,& 3π2 _ radians.

Explanation of Solution

Given information : Function f(x)=2sinx+cos2x ; Trigonometric Equation 2cosx−4sinxcosx=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 from left side of the equation 2cosx−4sinxcosx=0

2cosx(1−2sinx)=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.

2cosx=0▶1^s^t equation

1−2sinx=0▶2ⁿ^d equation

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

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

2cosx2=02cosx=0x=cos−1(0)x=π2, 3π2

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

By subtracting1 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

1−2sinx=01−2sinx−1=0−1−2sinx=−1−2sinx−2=−1−2sinx=12x=sin−1(12)x=π6, 5π6

Conclusion:

x=π2, 3π2, π6, & 5π6 radians are the zero of the function and solution to the equation 2cosx−4sinxcosx=0 in the interval [0,2π)

Chapter 5.3, Problem 81E

Chapter 5.3, Problem 83E

Chapter 5 Solutions

EBK PRECALCULUS W/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

Ch. 5.1 - Prob. 11E Ch. 5.1 - Prob. 12E Ch. 5.1 - Prob. 13E Ch. 5.1 - Prob. 14E Ch. 5.1 - Prob. 15E Ch. 5.1 - Prob. 16E Ch. 5.1 - Prob. 17E Ch. 5.1 - Prob. 18E Ch. 5.1 - Prob. 19E Ch. 5.1 - Prob. 20E Ch. 5.1 - Prob. 21E Ch. 5.1 - Prob. 22E Ch. 5.1 - Prob. 23E Ch. 5.1 - Prob. 24E Ch. 5.1 - Prob. 25E Ch. 5.1 - Prob. 26E Ch. 5.1 - Prob. 27E Ch. 5.1 - Prob. 28E Ch. 5.1 - Prob. 29E Ch. 5.1 - Prob. 30E Ch. 5.1 - Prob. 31E Ch. 5.1 - Prob. 32E Ch. 5.1 - Prob. 33E Ch. 5.1 - Prob. 34E Ch. 5.1 - Prob. 35E Ch. 5.1 - Prob. 36E Ch. 5.1 - Prob. 37E Ch. 5.1 - Prob. 38E Ch. 5.1 - Prob. 39E Ch. 5.1 - Prob. 40E Ch. 5.1 - Prob. 41E Ch. 5.1 - Prob. 42E Ch. 5.1 - Prob. 43E Ch. 5.1 - Prob. 44E Ch. 5.1 - Prob. 45E Ch. 5.1 - Prob. 46E Ch. 5.1 - Prob. 47E Ch. 5.1 - Prob. 48E Ch. 5.1 - Prob. 49E Ch. 5.1 - Prob. 50E Ch. 5.1 - Prob. 51E Ch. 5.1 - Prob. 52E Ch. 5.1 - Prob. 53E Ch. 5.1 - Prob. 54E Ch. 5.1 - Prob. 55E Ch. 5.1 - Prob. 56E Ch. 5.1 - Prob. 57E Ch. 5.1 - Prob. 58E Ch. 5.1 - Prob. 59E Ch. 5.1 - Prob. 60E Ch. 5.1 - 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