Problem 29 - Algebra and Trigonometry

Question

\(17-34\) . An equation is given. (a) Find all solutions of the equation. (b) Find the solutions in the interval \([0,2 \pi) .\) $$ \sec \theta-\tan \theta=\cos \theta $$

Step-by-Step Solution

Answer

The solutions in [0, 2π) are \(θ = 0, \frac{π}{2}, π\).

Step 1: Understand the Equation

We start by analyzing the given equation \(\sec \theta - \tan \theta = \cos \theta\). We recognize that \(\sec \theta = \frac{1}{\cos \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). The goal is to simplify and solve this equation.

Step 2: Simplify the Equation

Substitute the identities \(\sec \theta = \frac{1}{\cos \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) into the equation:\[\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \cos \theta.\]Combine the left-hand side into a single fraction:\[\frac{1 - \sin \theta}{\cos \theta} = \cos \theta.\]

Step 3: Clear the Fraction

Multiply both sides by \(\cos \theta\) to clear the fraction:\[1 - \sin \theta = \cos^2 \theta.\] Recall that \(\cos^2 \theta = 1 - \sin^2 \theta\). Substitute this into the equation:\[1 - \sin \theta = 1 - \sin^2 \theta.\]

Step 4: Rearrange and Solve

Rearrange the equation:\[\sin^2 \theta - \sin \theta = 0.\]Factor out \(\sin \theta\):\[\sin \theta (\sin \theta - 1) = 0.\]Set each factor to zero, giving us the solutions \(\sin \theta = 0\) and \(\sin \theta = 1\).

Step 5: Find General Solutions

Solve for \(\theta\):- For \(\sin \theta = 0\), \(\theta = n\pi\) where \(n\) is any integer.- For \(\sin \theta = 1\), \(\theta = \frac{\pi}{2} + 2k\pi\) where \(k\) is any integer.

Step 6: Find Solutions in Given Interval

Now find the solutions within the interval \([0, 2\pi)\):- From \(\sin \theta = 0\), \(\theta = 0, \pi\).- From \(\sin \theta = 1\), \(\theta = \frac{\pi}{2}\).

Step 7: Conclusion: Solutions in \([0, 2\pi)\)

The solutions within \([0, 2\pi)\) are:\[\theta = 0, \frac{\pi}{2}, \pi.\]

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in solving trigonometric equations. They help us rewrite trigonometric functions in different forms, enabling easier manipulation and simplification of equations. In the context of our exercise, we use two key identities:

• The secant identity: \[\sec \theta = \frac{1}{\cos \theta}\]
• The tangent identity: \[\tan \theta = \frac{\sin \theta}{\cos \theta}\]

By substituting these into the original equation \(\sec \theta - \tan \theta = \cos \theta\), the equation becomes easier to solve. These identities transform the equation into a rational form that can be simplified and manipulated to solve for \( \theta \). Using such identities allows us to express unfamiliar or complex trigonometric terms in a more recognizable manner, supporting clearer analysis and solution development.

Interval Solutions

Interval solutions refer to finding the values of \(\theta\) within a specified range. In trigonometry, it's common to restrict solutions to a standard interval, such as \([0, 2\pi)\). This is because trigonometric functions are periodic and can have multiple values, repeating every full cycle.
For this exercise, once we simplify and solve the equation, we find the general solutions:

• For \(\sin \theta = 0\), the general solution is \(\theta = n\pi\).
• For \(\sin \theta = 1\), the general solution is \(\theta = \frac{\pi}{2} + 2k\pi\).

However, we are only interested in solutions within \([0, 2\pi)\) in exercise part (b). Thus, evaluating for \(n\) and \(k\) within this interval gives us:

• From \(\sin \theta = 0\): \(\theta = 0, \pi\).
• From \(\sin \theta = 1\): \(\theta = \frac{\pi}{2}\).

Interval solutions allow us to identify specific answers within a desired range, tailoring general solutions to practical and applicable results.

Angle Measures

Angle measures from trigonometric solutions often require conversion or evaluation to make sense of them within defined intervals. Trigonometric equations typically cover an infinite number of solutions due to the periodic nature of trig functions.
Angles can be expressed in degrees or radians, with many trigonometric intervals using radians, as in this exercise with \([0, 2\pi)\).

• Radian measure is a natural unit for mathematicians, defined directly in the context of a circle where \(2\pi\) radians equals a full circle (360 degrees).
• Working on exercises in radians requires familiarity, as angles like 0, \(\pi/2\), and \(\pi\) correspond to specific positions on the unit circle (e.g., \(0\) radians being the positive \(x\)-axis).

Understanding and working with angle measures in radians enable us to efficiently determine and comprehend solutions to trigonometric problems in their most common context, making the process of finding valid solutions across desired intervals more intuitive.