Problem 107 - Algebra and Trigonometry

Question

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. $$ \cos x \csc x=2 \cos x $$

Step-by-Step Solution

Answer

The solutions to the equation are \( x = \frac{\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2} \)

Step 1: Replace \(\csc x\) with \(\frac{1}{\sin x}\)

The equation \(\cos x \csc x = 2 \cos x\) can be re-written as \( \cos x (\frac{1}{\sin x}) = 2 \cos x\) since \( \csc x \) is equivalent to \( \frac{1}{\sin x} \).

Step 2: Simplify the equation

Cancelling \( \cos x\) on both sides (provided \( \cos x \neq 0\)) we get \( \frac{1}{\sin x} = 2\). We cannot cancel out \( \cos x \) if \( x \) is \( \frac{\pi}{2} \) or \( \frac{3\pi}{2} \) because \( \cos(\frac{\pi}{2}) = \cos(\frac{3\pi}{2}) = 0\).

Step 3: Solve for \( x \)

To solve \( \frac{1}{\sin x} = 2 \) for \( x \), we flip both sides to get \( \sin x = \frac{1}{2} \). On the interval \( [0,2\pi) \), \( x \) is \( \frac{\pi}{6} \) or \( \frac{5\pi}{6} \), as those are the points where \( \sin x = \frac{1}{2} \).

Step 4: Account for excluded values

Now we must look back at step 2, where we excluded \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \) because they would result in division by zero. These are valid solutions to the original equation, because in those cases, both sides of the equation are undefined. Therefore they are included in the set of solutions.

Key Concepts

Cosecant Function

The cosecant function, written as \( \csc x \), is an essential trigonometric function. It's the reciprocal of the sine function, defined as \( \csc x = \frac{1}{\sin x} \). This function helps solve trigonometric equations by converting them into different forms.
By replacing \( \csc x \) with \( \frac{1}{\sin x} \), you can simplify equations that initially seem complex.
Keep in mind, for the cosecant function to be defined, \( \sin x \) cannot be zero. In such cases, \( x \) values where \( \sin x = 0 \) result in undefined conditions, which are crucial to recognize when solving equations.

Interval Notation

Interval notation is a way of representing the set of all possible values of a variable. In trigonometry, it often indicates the domain within which solutions are sought.
The given problem specifies the interval \([0, 2\pi)\), meaning solutions should be found between 0 and just under \( 2\pi \). This interval is crucial because it helps pinpoint specific angles where trigonometric functions behave in particular ways, aiding in the identification of exact solutions.

Solving Trigonometric Equations

Solving trigonometric equations involves finding the angles that make the equation true. This often requires simplifying the equation first.
For example, by substituting \( \csc x \) with \( \frac{1}{\sin x} \), our equation can be manipulated into a simpler form: \( \frac{1}{\sin x} = 2 \), which then simplifies to \( \sin x = \frac{1}{2} \).
It is also crucial to check if any simplifications result in conditions where the original equation is undefined, as these values might need special attention.

Exact Trigonometric Values

Exact trigonometric values refer to known angles where the sine, cosine, and other trigonometric functions have specific, precise values.
For \( \sin x = \frac{1}{2} \), we know from trigonometric tables or the unit circle that \( x \) can be \( \frac{\pi}{6} \) or \( \frac{5\pi}{6} \) within the interval \([0, 2\pi)\).
Recognizing and using these exact values helps avoid gradation to decimal approximations, aiding in clean and precise solutions.