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For Exercises \(51-54,\) solve for the angle \(\theta,\) where \(0 \leq \theta \leq 2 \pi\) $$\cos 2 \theta+\cos \theta=0$$

Short Answer

Expert verified
\(\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \pi\) in the interval \(0 \leq \theta \leq 2\pi\).

Step by step solution

01

Rewrite Equation

The given equation is \( \cos 2\theta + \cos \theta = 0 \). Use the identity \( \cos 2\theta = 2\cos^2 \theta - 1 \) to rewrite the fraction as \( 2\cos^2 \theta - 1 + \cos \theta = 0 \).
02

Form a Quadratic Equation

Rearrange the equation to obtain a standard quadratic form: \( 2\cos^2 \theta + \cos \theta - 1 = 0 \).
03

Use the Substitution

Let \( x = \cos \theta \). The equation becomes \( 2x^2 + x - 1 = 0 \). This is a quadratic equation in terms of \( x \).
04

Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = 1, c = -1 \). Substitute these values into the formula to get \( x = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-1)}}{2 \times 2} \), which simplifies to \( x = \frac{-1 \pm \sqrt{9}}{4} \).
05

Calculate the Roots

This gives the solutions \( x = \frac{-1 + 3}{4} \) and \( x = \frac{-1 - 3}{4} \), which are \( x = \frac{1}{2} \) and \( x = -1 \).
06

Find \(\theta\) for Each \(x\) Value

1. For \( x = \frac{1}{2} \), \( \cos \theta = \frac{1}{2} \). The angles that satisfy this in the interval \( 0 \leq \theta \leq 2\pi \) are \( \theta = \frac{\pi}{3} \) and \( \theta = \frac{5\pi}{3} \).2. For \( x = -1 \), \( \cos \theta = -1 \). The angle that satisfies this is \( \theta = \pi \).
07

Combine All Solutions

The complete set of solutions for \( \theta \) in the specified interval is \( \theta = \frac{\pi}{3}, \frac{5\pi}{3}, \) and \( \pi \).

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Key Concepts

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

Cosine Function
The cosine function, denoted as \( \cos \theta \), is one of the fundamental trigonometric functions. It's primarily used to determine the horizontal coordinate of a point on the unit circle. Understanding cosine is crucial as it helps solve various trigonometric equations and problems. For an angle \( \theta \), \( \cos \theta \) gives the x-coordinate of the corresponding point on the unit circle.

Key properties of the cosine function include:
  • Range: The values of cosine range from -1 to 1.
  • Periodicity: The cosine function is periodic with a period of \( 2\pi \). This means that \( \cos(\theta) = \cos(\theta + 2\pi n) \) for any integer \( n \).
  • Symmetry: The function is even, meaning \( \cos(-\theta) = \cos(\theta) \).
These characteristics help in identifying solutions within specific intervals, such as between 0 and \( 2\pi \), as seen in the given problem. By knowing these angles, we can identify particular solutions that meet the equation’s criteria.
Quadratic Equations
In trigonometry, quadratic equations frequently appear, often hidden within trigonometric functions like cosine. A quadratic equation is generally in the form \( ax^2 + bx + c = 0 \). Solving them can yield the specific values needed for additional calculations.

To solve a quadratic equation, there are several methods, with the quadratic formula being among the most reliable:
  • Quadratic Formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
  • Factoring: If easily factorable, you can rewrite the equation to find its roots directly.
  • Completing the Square: A method where the equation is rearranged into a perfect square trinomial.
For cosine-related equations, substitution is often used to transform a trigonometric expression into a quadratic equation, making it easier to solve. In the exercise, \( \cos \theta \) was substituted with \( x \) to apply the quadratic formula, solving for the cosine values that satisfy the original equation.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for any angle. These identities simplify and solve complex trigonometric equations. Knowing and applying these can vastly simplify problem-solving.

Here are critical identities relevant to understanding trigonometric solutions:
  • Double Angle Identity: \( \cos 2\theta = 2\cos^2 \theta - 1 \) helps rewrite terms involving \( \cos 2\theta \) into useful quadratic expressions.
  • Pythagorean Identity: \( \cos^2 \theta + \sin^2 \theta = 1 \). This relates the cosine and sine functions.
Using these identities, like the double angle identity in this exercise, turns challenging expressions into manageable quadratic equations. Consequently, it becomes straightforward to apply algebraic techniques for solutions. Recognizing when and how to use these identities is a fundamental skill in trigonometry.

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Most popular questions from this chapter

Exercises \(27-36\) tell how many units and in what directions the graphs of the given equations are to be shifted. Give an equation for the shifted graph. Then sketch the original and shifted graphs together, labeling each graph with its equation. $$ y=\frac{1}{2}(x+1)+5 \quad \text { Down } 5, \text { right } 1 $$

Use the addition formulas to derive the identities in Exercises \(31-36\) $$\sin (A-B)=\sin A \cos B-\cos A \sin B$$

In Exercises \(71-74,\) you will explore graphically the general sine function $$f(x)=A \sin \left(\frac{2 \pi}{B}(x-C)\right)+D$$ as you change the values of the constants \(A, B, C,\) and \(D .\) Use a CAS or computer grapher to perform the steps in the exercises. The amplitude \(A\) Set the constants \(B=6, C=D=0\) . a. Describe what happens to the graph of the general sine function as \(A\) increases through positive values. Confirm your answer by plotting \(f(x)\) for the values \(A=1,5,\) and \(9 .\) b. What happens to the graph for negative values of \(A ?\)

Use graphing software to graph the functions specified in Exercises \(31-36\) . Select a viewing window that reveals the key features of the function. Graph two periods of the function \(f(x)=3 \cot \frac{x}{2}+1\)

Industrial costs A power plant sits next to a river where the river is 800 ft wide. To lay a new cable from the plant to a location in the city 2 mi downstream on the opposite side costs \(\$ 180\) per foot across the river and \(\$ 100\) per foot along the land. a. Suppose that the cable goes from the plant to a point \(Q\) on the opposite side that is \(x\) ft from the point \(P\) directly opposite the plant. Write a function \(C(x)\) that gives the cost of laying the cable in terms of the distance \(x\) . b. Generate a table of values to determine if the least expensive location for point \(Q\) is less than 2000 ft or greater than 2000 ft from point \(P .\)

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