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Solve the following equations. $$2 \theta \cos \theta+\theta=0$$

Short Answer

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Question: Solve the trigonometric equation $$2 \theta \cos \theta+\theta=0$$ and state the general solution. Answer: The general solution of the given equation is: $$\theta=0$$ or $$\theta=\arccos\left(-\frac{1}{2}\right)+2n\pi$$ or $$\theta=-\arccos\left(-\frac{1}{2}\right)+2n\pi$$ Where \(n\) is an integer.

Step by step solution

01

Factor out \(\theta\) from the equation

We can factor out \(\theta\) from the equation to simplify it: $$\theta(2 \cos \theta + 1) = 0$$
02

Solve for \(\theta\) in two cases

Now, we have a product equal to 0, which means that either \(\theta=0\) or the other factor is equal to 0, i.e., \(2\cos\theta+1=0\). Let's solve these two cases separately.
03

Solve the first case: \(\theta=0\)

From the factoring step, we already know that \(\theta=0\) is a solution to the equation.
04

Solve the second case: \(2\cos\theta+1=0\)

To solve the equation \(2\cos\theta+1=0\), we need to isolate \(\cos\theta\): $$\cos\theta=-\frac{1}{2}$$
05

Find the general solution for \(2\cos\theta+1=0\)

We know that $$\cos\theta=-\frac{1}{2}$$ and need to find the general solution of this equation. The inverse cosine function, \(\arccos\), can be used to find the angles that have a cosine equal to \(-\frac{1}{2}\): $$\theta=\arccos\left(-\frac{1}{2}\right)+2n\pi$$ or $$\theta=-\arccos\left(-\frac{1}{2}\right)+2n\pi$$ Where \(n\) is an integer.
06

State the general solution of the given equation

Combining the solutions from Steps 3 and 5, the general solution of the given equation $$2 \theta \cos \theta + \theta = 0$$ is: $$\theta=0$$ or $$\theta=\arccos\left(-\frac{1}{2}\right)+2n\pi$$ or $$\theta=-\arccos\left(-\frac{1}{2}\right)+2n\pi$$ Where \(n\) is an integer.

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Key Concepts

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

Factoring
Factoring is a technique used to simplify equations and make them easier to solve. In this problem, the equation is initially given as \(2 \theta \cos \theta + \theta = 0\). Notice that both terms in this equation have a common factor, \(\theta\). Factoring out \(\theta\) from the expression, we rewrite the equation as \(\theta(2 \cos \theta + 1) = 0\).

This transformation allows us to address each part separately. The factored form is crucial because it breaks down the equation into simpler components. Once factored, the properties of zero come into play, simplifying the solving process.
General Solution
A general solution in trigonometric equations refers to all possible solutions which satisfy the equation. This is important because trigonometric functions are periodic, and hence have multiple solutions within any given interval.

For this problem, once the equation \(\theta(2 \cos \theta + 1) = 0\) was factored, it resulted in two separate equations to solve independently: \(\theta = 0\) and \(2 \cos \theta + 1 = 0\).

The solution \(\theta = 0\) is straightforward, while the equation \(2 \cos \theta + 1 = 0\) requires a few more steps. For it, the general solutions are described by the formulae \(\theta = \arccos\left(-\frac{1}{2}\right) + 2n\pi\) and \(\theta = -\arccos\left(-\frac{1}{2}\right) + 2n\pi\), where \(n\) is any integer.

This approach ensures that all of the infinite number of solutions are expressed.
Inverse Cosine
The inverse cosine function, denoted as \(\arccos\), is a fundamental concept used to find angles when the cosine value is known. In this exercise, we ended up with the equation \(\cos \theta = -\frac{1}{2}\).

Using the \(\arccos\) function, we determine that \(\theta = \arccos\left(-\frac{1}{2}\right)\). The use of \(\arccos\) lets us retrieve the principal angle which has the given cosine value.

However, angles with the same cosine value recur in cycles. That's why the general solution also adds multiples of \(2\pi\), representing one full cycle of a cosine wave, to the principal angle, thereby capturing all possible solutions. This cyclic property underlies trigonometric functions' periodic nature.
Product Equal to Zero
A product is equal to zero when at least one of its factors equals zero. This is known as the Zero Product Property. In the factored equation \(\theta(2 \cos \theta + 1) = 0\), we use this property to divide the problem into manageable pieces.

This means either \(\theta = 0\) or \(2 \cos \theta + 1 = 0\). Breaking it down this way turns the problem into simpler parts that can each be solved on their own.

Recognizing such properties is extremely useful in algebra and trigonometry. It simplifies not only trigonometric equations but is also applicable in various algebraic situations, making problem-solving more efficient and less error-prone.

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Most popular questions from this chapter

A GPS device tracks the elevation \(E\) (in feet) of a hiker walking in the mountains. The elevation \(t\) hours after beginning the hike is given in the figure. a. Find the slope of the secant line that passes through points \(A\) and \(B\). Interpret your answer as an average rate of change over the interval \(1 \leq t \leq 3\) b. Repeat the procedure outlined in part (a) for the secant line that passes through points \(P\) and \(Q\) c. Notice that the curve in the figure is horizontal for an interval of time near \(t=5.5 \mathrm{hr}\). Give a plausible explanation for the horizontal line segment. (GRAPH CAN'T COPY)

Daylight function for \(40^{\circ} \mathrm{N}\) Verify that the function $$D(t)=2.8 \sin \left(\frac{2 \pi}{365}(t-81)\right)+12$$ has the following properties, where \(t\) is measured in days and \(D\) is the number of hours between sunrise and sunset. a. It has a period of 365 days. b. Its maximum and minimum values are 14.8 and \(9.2,\) respectively, which occur approximately at \(t=172\) and \(t=355\) respectively (corresponding to the solstices). c. \(D(81)=12\) and \(D(264) \approx 12\) (corresponding to the equinoxes).

Shifting and scaling Use shifts and scalings to graph the given functions. Then check your work with a graphing utility. Be sure to identify an original function on which the shifts and scalings are performed. \(f(x)=x^{2}-2 x+3\) (Hint: Complete the square first.)

Without using a graphing utility, sketch the graph of \(y=\log _{2} x .\) Then on the same set of axes, sketch the graphs of \(y=\log _{2}(x-1), y=\log _{2} x^{2}\) \(y=\left(\log _{2} x\right)^{2},\) and \(y=\log _{2} x+1\)

Shifting and scaling Use shifts and scalings to graph the given functions. Then check your work with a graphing utility. Be sure to identify an original function on which the shifts and scalings are performed. $$g(x)=-3 x^{2}$$

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