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Round approximate answers in radians to four decimal places and approximate answers in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$\cot x+\sqrt{3}=0$$

Short Answer

Expert verified
The least nonnegative angle that satisfies \( \cot x + \sqrt{3} = 0 \) is \( x = \frac{5\pi}{6} \).

Step by step solution

01

Rewrite the Equation

First, we need to rewrite the original equation. The given equation is \( \cot x + \sqrt{3} = 0 \). We can rewrite this as \( \cot x = -\sqrt{3} \). Our goal is to find the value of \( x \) that satisfies this equation.
02

Convert Cotangent to Tangent

Recall that \( \cot x = \frac{1}{\tan x} \). Thus, \( \cot x = -\sqrt{3} \) implies \( \frac{1}{\tan x} = -\sqrt{3} \), or equivalently \( \tan x = -\frac{1}{\sqrt{3}} \).
03

Identify Reference Angle in Radians

The value \( \tan x = -\frac{1}{\sqrt{3}} \) corresponds to the angles where tangent is \( \frac{1}{\sqrt{3}} \), which are \( x = \frac{\pi}{6} \) and other angles that satisfy the equation. However, since we need the angle in tangent's negative range, we look at the specific quadrants where tangent is negative.
04

Determine Least Nonnegative Angle in Radians

Tangent is negative in the second and fourth quadrants. The reference angle is \( \frac{\pi}{6} \), so the angles are \( x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \) and \( x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \).

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Key Concepts

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

Radian Measure
When it comes to angles, degrees are perhaps the most familiar unit; however, in trigonometry, radians are the preferred measure for angle calculations. A radian is based on the radius of a circle. Specifically, one radian is the angle made when the arc length is equal to the radius of the circle. This relationship is key because it ties directly into the geometry of a circle, making calculations more natural in mathematical analysis.
In the solution, when angles appear as multiples of \( \pi \), they're measured in radians. For instance, \( \pi \) radians is equivalent to 180 degrees, forming a direct link between the two systems. To find solutions in the least nonnegative angle, we often work within the range from 0 to \( 2\pi \). This range describes a full circle, where starting angles reset after \( 2\pi \) radians.
Radians are effective as they allow for more streamlined calculus operations and are essential when working with periodic functions like sine, cosine, and tangent seen in trigonometry.
Reference Angle
A reference angle is a simplified version of an angle, stripped of its direction in a unit circle, making calculations easier. The reference angle of any angle is essentially the smallest angle between the terminal side of the original angle and the horizontal axis. When it comes to solving trigonometric equations, the reference angle helps determine where the given angle's trigonometric function might take on positive or negative values.
Consider the function \( \tan x = -\frac{1}{\sqrt{3}} \). The equation implies that the tangent corresponds to specific angles where the tangent value would be \( \frac{1}{\sqrt{3}} \) referenced to either the second or fourth quadrant, where the tangent is negative. Hence, the reference angle of \( \frac{\pi}{6} \) lets us identify the key angles \( \frac{5\pi}{6} \) and \( \frac{11\pi}{6} \), which lie in these specific quadrants.
Cotangent Function
The cotangent function, denoted as \( \cot x \), is the reciprocal of the tangent function. Simply put, if the tangent of an angle is defined as the ratio of the opposite side to the adjacent side in a right-angled triangle, the cotangent is the ratio of the adjacent side to the opposite side.
Mathematically, this means \( \cot x = \frac{1}{\tan x} \). This equality is pivotal in transforming a problem involving cotangent to one involving tangent, which can be more straightforward to handle, especially when employing angle measures like radians.
For the equation \( \cot x = -\sqrt{3} \), it is equivalent to \( \tan x = -\frac{1}{\sqrt{3}} \), thereby enabling us to consider the angles at which this specific tangent value as achieved. Recognizing these transformations enhances the ease of solving trigonometric equations.
Tangent Function
The tangent function, \( \tan x \), is a fundamental trigonometric function. It expresses the ratio of the opposite side to the adjacent side in a right-angled triangle. Unlike sine and cosine functions, the tangent function can take any real number value, ranging from negative infinity to positive infinity.
In solving equations like \( \tan x = -\frac{1}{\sqrt{3}} \), understanding the behavior of the tangent function is crucial. The tangent function is periodic with a period of \( \pi \), meaning its values repeat every \( \pi \) radians. This periodicity helps in finding all possible solutions to the equation within a given range, like \([0, 2\pi) \).
Knowing that the function is negative in the second and fourth quadrants provides insight into solving for the angles. Thus, the respective angles are \( \frac{5\pi}{6} \) and \( \frac{11\pi}{6} \), using the reference angle \( \frac{\pi}{6} \) to establish these angles' exact positions.

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Most popular questions from this chapter

Use a graphical method to solve each equation over the interval \([0,2 \pi) .\) Round values to the nearest thousandth. $$\sin \frac{x}{2}+\cos 3 x=0$$

Sound Waves Sound is a result of waves applying pressure to a person's eardrum. For a pure sound wave radiating outward in a spherical shape, the trigonometric function $$P=\frac{a}{r} \cos \left(\frac{2 \pi r}{\lambda}-c t\right)$$ can be used to model the sound pressure \(P\) at a radius of \(r\) feet from the source, where \(t\) is time in seconds, \(\lambda\) is length of the sound wave in feet, \(c\) is speed of sound in feet per second, and \(a\) is maximum sound pressure at the source measured in pounds per square foot. (Source: Beranek, L.., Noise and Vibration Control, Institute of Noise Control Engineering. Washington, DC.) Let \(\lambda=4.9\) feet and \(c=1026\) feet per second. (IMAGE CANNOT COPY) (a) Let \(a=0.4\) pound per square foot. Graph the sound pressure at a distance \(r=10\) feet from its source over the interval \(0 \leq t \leq 0.05 .\) Describe \(P\) at this distance. (b) Now let \(a=3\) and \(t=10 .\) Graph the sound pressure for \(0 \leq r \leq 20 .\) What happens to the pressure \(P\) as the radius \(r\) increases? (c) Suppose a person stands at a radius \(r\) so that $$r=n \lambda$$ where \(n\) is a positive integer. Use the difference identity for cosine to simplify \(P\) in this situation.

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Solve each problem. Hearing Beats in Music Musicians sometimes tune instruments by playing the same tone on two different instruments and listening for a phenomenon known as beats. When the two instruments are in tune, the beats disappear. The ear hears beats because the pressure slowly rises and falls as a result of the slight variation in the frequency. This phenomenon can be seen on a graphing calculator. (a) Consider two tones with frequencies of 220 and \(223 \mathrm{Hz}\) and pressures \(P_{1}=0.005 \sin 440 \pi t \quad\) and \(\quad P_{2}=0.005 \sin 446 \pi t\) respectively. A graph of \(P_{1}+P_{2}\) as \(Y_{3}\) felt by an eardrum over the 1 -second interval \([0.15,1.15]\) is shown here. How many beats are there in 1 second? (Graph can't copy) (b) Repeat part (a) with frequencies of 220 and 216 (c) Determine a simple way to find the number of beats per second if the frequency of each tone is given.

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