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Use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$2 \cos 2 x+1=0$$

Short Answer

Expert verified
The solutions for the given equation within the interval [0, \(2\pi\)] are \(x = \pi/3\) and \(x = 2\pi/3\).

Step by step solution

01

Set the problem

The given equation is \(2 \cos 2x + 1 = 0\). The goal is to find \(x\) such that the equation holds true. First, move 1 to the right hand side of the equation.
02

Solve for the function

Now, the equation becomes \(2 \cos 2x = -1\). After that, divide both sides by 2 to solve for cosine function, you get \(\cos 2x = -1鈦2\).
03

Use the inverse cosine function

For usual cosine function, the answer varies between 0 and 180 degree or 0 and \(\pi\). However, for \(\cos 2x\), the interval will be from 0 to 360 degrees or from 0 to \(2\pi\). Cosine value being -1/2 corresponds to \(2\pi/3\) and \(4\pi/3\) in the interval [0, \(2\pi\)]. Thus, \(2x = 2\pi/3\) or \(2x = 4\pi/3\).
04

Solve for \(x\)

Finally, dividing both sides of the equations above by 2, the solutions for \(x\) within the interval [0, \(2\pi\)] are \(x = \pi/3\) or \(x = 2\pi/3\).

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Key Concepts

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

Cosine Function
Understanding the cosine function is crucial when dealing with trigonometric equations. In essence, the cosine function relates the angle of a right triangle to the ratio of the adjacent side to the hypotenuse. It's one of the basic trigonometric functions and is periodic, meaning it repeats its values in a predictable pattern.

When solving equations like the given exercise, it is important to remember the cosine function's periodicity. It has a period of \(2\pi\) radians, meaning that it repeats every \(2\pi\) radians. This is why, when finding solutions to \(\cos(2x) = -\frac{1}{2}\), we look for all angles within the interval \( [0, 2\pi) \) that satisfy the equation.

Keep in mind that the cosine function also has particular values for certain angles, which are often referred to as 'cosine of special angles'. These include angles like 0, \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\), and \(\frac{\pi}{2}\), where the cosine values are 1, \(\sqrt{3}/2\), \(\sqrt{2}/2\), 1/2, and 0 respectively. Knowing these can help solve trigonometric equations more efficiently.
Inverse Trigonometric Functions
When we have an equation involving a trigonometric function and we need to isolate the variable, we use an inverse trigonometric function. They are the inverses of the standard trigonometric functions and are used to find the angle if the trigonometric ratio is known.

In our example \(2 \cos 2x + 1 = 0\), after isolating the cosine function, we apply the inverse cosine, denoted as \(\cos^{-1}\) or \(\arccos\), to both sides. This will give us the angles that correspond to the value \(\cos(2x) = -\frac{1}{2}\).

These functions, however, have a limited range. For the inverse cosine function, the principal values lie in the interval \( [0, \pi] \) for radians. Be cautious when dealing with other intervals; make sure you consider the periodic nature of the trigonometric functions to find all possible solutions.
Radians and Degrees
Radians and degrees are two units for measuring angles, and understanding both is fundamental for solving trigonometric equations effectively. A full circle is measured as \(360^\circ\) in degrees and \(2\pi\) radians. Thus, for conversions, remember that \(\pi\) radians is equivalent to \(180^\circ\).

In our exercise, we're working within the interval \( [0, 2\pi) \) radians. This means we are considering all angles that start from 0 up to but not including a full circle. Performing trigonometric functions in radians is often more natural in calculus and higher mathematics because radian measures are directly related to the lengths of arcs in a circle.

When you encounter trigonometric equations, it might be necessary to convert degrees to radians or vice versa to use special angle values or to apply certain calculus principles. Never forget to perform this conversion when it's needed, as this ensures proper interpretation and solution of the problems.

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Most popular questions from this chapter

Find the inverse of \(f(x)=\frac{x-1}{x+1}\) (Section \(1.8, \text { Example } 4)\)

Will help you prepare for the material covered in the next section. Use the appropriate values from Exercise 101 to answer each of the following. a. Is \(\sin \left(30^{\circ}+60^{\circ}\right),\) or \(\sin 90^{\circ},\) equal to \(\sin 30^{\circ}+\sin 60^{\circ} ?\) b. Is \(\sin \left(30^{\circ}+60^{\circ}\right),\) or \(\sin 90^{\circ},\) equal to \(\sin 30^{\circ} \cos 60^{\circ}+\cos 30^{\circ} \sin 60^{\circ} ?\)

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. $$\sin x=-\cos x \tan (-x)$$

Use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$2 \tan ^{2} x+5 \tan x+3=0$$

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.

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