Problem 24 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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 \( \frac{1}{\sin^2 x} = 2 \frac{\cos x}{\sin x} \).

Step 2: Clear the Denominator

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

Step 3: 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 \( 1 = \sin(2x) \).

Step 4: 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, \( 2x = \frac{\pi}{2} + 2k\pi \).

Step 5: 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 \( x = \frac{5\pi}{4} \).

Key Concepts

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.