/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Solve each equation for solution... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve each equation for solutions over the interval \([0,2 \pi)\) by first solving for the trigonometric finction. Do not use a calculator. $$\csc ^{2} x=2 \cot x$$

Short Answer

Expert verified
The solutions are \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).

Step by step solution

01

Rewrite the Equation in Terms of Sine and Cosine

The given equation is \( \csc^2 x = 2 \cot x \). We can rewrite this using the identities \( \csc x = \frac{1}{\sin x} \) and \( \cot x = \frac{\cos x}{\sin x} \). Therefore, \( \csc^2 x = \frac{1}{\sin^2 x} \) and \( \cot x = \frac{\cos x}{\sin x} \). Substitute these to get \( \frac{1}{\sin^2 x} = 2 \frac{\cos x}{\sin x} \).
02

Clear the Denominator

Multiply both sides of the equation by \( \sin^2 x \) to eliminate the fraction: \( 1 = 2 \cos x \sin x \).
03

Simplify Using Trigonometric Identities

Use the double angle identity, which states that \( \sin(2x) = 2\sin x \cos x \). Substituting this into the equation gives \( 1 = \sin(2x) \).
04

Solve the Simplified Equation

Find the values of \( 2x \) such that \( \sin(2x) = 1 \). We know that \( \sin \theta = 1 \) when \( \theta = \frac{\pi}{2} + 2k\pi \), where \( k \) is an integer. Thus, \( 2x = \frac{\pi}{2} + 2k\pi \).
05

Solve for \( x \) and Adjust to Interval

Divide the equation by 2 to solve for \( x \): \( x = \frac{\pi}{4} + k\pi \). To find solutions within the interval \([0, 2\pi)\), consider \( k = 0 \) and \( k = 1 \). This gives solutions \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).

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Key Concepts

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

Trigonometric Identities
Trigonometric identities are fundamental to understanding and solving trigonometric equations. They allow us to rewrite complex expressions in simpler forms, making them easier to solve. Two important identities used in the exercise are the reciprocal and the ratio identities:
  • Reciprocal Identity for Cosecant: \( \csc x = \frac{1}{\sin x} \)
  • Reciprocal Identity for Cotangent: \( \cot x = \frac{\cos x}{\sin x} \)
These identities are used to transform the original equation \( \csc^2 x = 2 \cot x \) into an equation involving only sine and cosine. By expressing cosecant and cotangent in terms of sine and cosine, we simplify the equation to a form \( \frac{1}{\sin^2 x} = 2 \frac{\cos x}{\sin x} \). By integrating these identities effectively, you can tackle various trigonometric equations much more effortlessly.
Solving Equations
The art of solving equations revolves around isolating the variable in question. Once we have expressed the equation in terms of sines and cosines, the next step involves simplifying and solving for the desired trigonometric function. In our example, the equation \( 1 = 2 \cos x \sin x \) is simplified using the double angle formula:
  • Double Angle Identity: \( \sin(2x) = 2 \sin x \cos x \)
By recognizing that \( 2 \sin x \cos x \) equals \( \sin(2x) \), we replace it in the equation to yield \( 1 = \sin(2x) \). Solving this in the typical manner, we look for angles \( 2x \) where sine equals 1, specifically at \( \frac{\pi}{2} \). These solutions are derived by equating \( 2x = \frac{\pi}{2} + 2k\pi \), where \( k \) is an integer. From there, solving for \( x \) involves dividing by 2, providing candidate solutions sighted in the specific interval.
Angles in Radians
Radians are a unit of angular measure used in mathematics, specifically in trigonometry. Unlike degrees, which divide a circle into 360 parts, radians are based on the radius of the circle. One complete revolution around a circle is \( 2\pi \) radians. Therefore, common angles often used are:
  • \( \frac{\pi}{2} \) radians (90 degrees)
  • \( \pi \) radians (180 degrees)
  • \( 2\pi \) radians (360 degrees)
In the context of the problem, the solution for \( x \) is presented in radians between \( 0 \) and \( 2\pi \). It necessitates familiarity with the radian measure so that trigonometric solutions align within the given interval. Understanding radians is essential when navigating these equations, as it is the primary unit used in calculus and higher-level mathematics.
Interval Notation
Interval notation is a way to express the set of solutions or domain of a function within certain boundaries. In our exercise, the solution is confined to the interval \([0, 2\pi)\). This notation means that we consider all values of \( x \) that are greater than or equal to 0 but less than \( 2\pi \).
  • Brackets: \([ \ ]\) denote inclusion of an endpoint, whereas \(( \ \)) denote exclusion.
  • For \([0, 2\pi)\), 0 is included while \(2\pi\) is not. This denotes a half-open interval.
By defining intervals this way, students can clearly see the limits within which to find the solution for \( x \). In our case, which yielded solutions \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \), ensures clarity in which values of \( x \) fit within the prescribed range.

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Most popular questions from this chapter

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.

Use a calculator to find each value. $$\cot (\arccos 0.58236841)$$

Solve each problem. A painting 3 feet high and 6 feet from the floor will cut off an angle $$\theta=\tan ^{-1}\left(\frac{3 x}{x^{2}+4}\right)$$ to an observer. Assume that the observer is \(x\) feet from the wall where the painting is displayed and that the eyes of the observer are 5 feet above the ground. (IMAGE CAN'T COPY). Find the value of \(\theta\) for each value of \(x\) to the nearest degree. (a) \(x=3\) (b) \(x=6\) (c) \(x=9\) (d) Derive the given formula for \(\theta\). (Hint: Use right triangles and the identity for \(\tan (\theta-\alpha) .)\) (e) Graph the function for \(\theta\) with a calculator, and determine the distance that maximizes the angle.

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

Solve each equation ( \(x\) in radians and \(\theta\) in degrees) for all exact solutions where appropriate. Round approximale values in radians to four decimal places and approximate values in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$-2 \cos 2 x=\sqrt{3}$$

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