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Chapter 5: Problem 51

Involve trigonometric equations quadratic in form. Solve each equation on the interval \([0,2 \pi)\) $$\sec ^{2} x-2=0$$

Short Answer

Expert verified
The solutions to the equation on the interval \([0, 2\pi)\) are \(x = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4\).

Step by step solution

01

Rewrite the Equation

The secant function is the reciprocal of the cosine function, meaning \(sec(x) = 1/cos(x)\). Therefore, the equation \(sec^2(x) - 2 = 0\) can be rewritten as \((1/cos^2(x)) - 2 = 0\).
02

Solve the Equation

Rearrange the equation from step 1 to get \(cos^2(x) = 1/2\). Then, take the square root of both sides to get \(cos(x) = \pm sqrt{1/2}\). In terms of \(\pi\), we have \(cos(x) = \pm sqrt{1/2} = \pm 1/sqrt{2} = \pm sqrt{2}/2\). Therefore, \(x = \pm \pi/4, \pm 3\pi/4, \pm 5\pi/4, \pm 7\pi/4\). However, because the range is from 0 to \(2\pi\), discard the negative solutions. Therefore, \(x = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4\).
03

Check Solutions

For each of the values of \(x\), check that it is a valid solution to the original equation \(sec^2(x) - 2 = 0\) by substituting \(x\) back into the equation and verify.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Form
Quadratic forms appear in trigonometric equations when dealing with squared functions. In this exercise, the given equation is \( \sec^2(x) - 2 = 0 \), which resembles the typical quadratic form \( ax^2 + bx + c = 0 \). The presence of \( \sec^2(x) \) indicates the equation is quadratic in nature.

Here's how this resemblance helps in solving the equation:
  • Just like we can solve quadratic equations by factoring, completing the square, or using the quadratic formula, similar strategies can apply here to simplify and solve.
  • Rearranging \( \sec^2(x) - 2 = 0 \) as \( \sec^2(x) = 2 \) aligns with isolating the squared term in a quadratic equation, allowing us to find solutions based on specific trigonometric identities.
Recognizing these trigonometric identities as quadratic forms allows for easier manipulation and solution finding.
Secant Function
The secant function, denoted as \( \sec(x) \), relates to the cosine function as its reciprocal, meaning \( \sec(x) = \frac{1}{\cos(x)} \). Here are a few important points about the secant function:

  • The secant function is defined for values where \( \cos(x) eq 0 \), and it inherits the periodicity of \( \cos(x) \) with a period of \( 2\pi \).
  • In this problem, converting \( \sec(x) \) to \( \frac{1}{\cos(x)} \) directly impacts the equation, allowing it to transform into a more familiar form: \( \frac{1}{\cos^2(x)} - 2 = 0 \).
Understanding secant as a reciprocal function makes it easier to relate to other trigonometric identities, enabling substituting between secant and cosine.
Cosine Function
The cosine function, \( \cos(x) \), is crucial in solving the equation \( \sec^2(x) - 2 = 0 \). By expressing \( \sec^2(x) \) as \( \frac{1}{\cos^2(x)} \), the equation becomes solvable through cosine. Here鈥檚 why cosine is important:

  • Cosine values can be directly connected to basic angles known from the unit circle, especially those associated with multiples of \( \pi/4 \) where \( \cos(x) = \frac{\sqrt{2}}{2} \). In this exercise, solutions are derived from recognizing these specific angles.
  • Transforming the original equation to \( \cos^2(x) = \frac{1}{2} \) and solving \( \cos(x) = \pm \frac{\sqrt{2}}{2} \) leads to identifying the angles \( x = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4 \) within the interval \( [0, 2\pi) \).
Grasping the role of the cosine function within this context not only simplifies the equation but also assists in determining the correct solutions efficiently.

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Most popular questions from this chapter

Write each trigonometric expression as an algebraic expression (that is, without any trigonometric fienctions). Assume that \(x\) and \(y\) are positive and in the domain of the given inverse trigonometric function. $$\cos \left(\sin ^{-1} x-\cos ^{-1} y\right)$$

Graph each side of the equation in the same viewing rectangle. If the graphs appear to coincide, verify that the equation is an identity. If the graphs do not appear to coincide, this indicates the equation is not an identity. In these exercises, find a value of \(x\) for which both sides are defined but not equal. $$\frac{\sin x}{1-\cos ^{2} x}=\csc x$$

A city's tall buildings and narrow streets reduce the amount of sunlight. If \(h\) is the average height of the buildings and \(w\) is the width of the street, the angle of elevation from the street to the top of the buildings is given by the trigonometric equation $$\tan \theta=\frac{h}{w}$$ A value of \(\theta=63^{\circ}\) can result in an \(85 \%\) loss of illumination. Some people experience depression with loss of sunlight. Determine whether such a person should live on a city street that is 80 feet wide with buildings whose heights average 400 feet. Explain your answer and include \(\theta,\) to the nearest degree, in your argument.

Will help you prepare for the material covered in the first section of the next chapter. Solve each equation by using the cross-products principle to clear fractions from the proportion: If \(\frac{a}{b}=\frac{c}{d},\) then \(a d=b c,(b \neq 0 \text { and } d \neq 0)\) Round to the nearest tenth. $$\text { Solve for } B: \frac{51}{\sin 75^{\circ}}=\frac{71}{\sin B}$$

Exercises \(116-118\) will help you prepare for the material covered in the next section. In each exercise, use exact values of trigonometric functions to show that the statement is true. Notice that each statement expresses the product of sines and/or cosines as a sum or a difference. $$\sin \pi \cos \frac{\pi}{2}=\frac{1}{2}\left[\sin \left(\pi+\frac{\pi}{2}\right)+\sin \left(\pi-\frac{\pi}{2}\right)\right]$$
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