Problem 34 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation for solutions over the interval \(\left[0^{\circ}, 360^{\circ}\right) .\) Give solutions to the nearest tenth as appropriate. $$\sin ^{2} \theta \cos \theta=\cos \theta$$

Step-by-Step Solution

Answer

The solutions are \(90^{\circ}\) and \(270^{\circ}\).

Step 1: Factor the Equation

Start with the given equation \( \sin^2 \theta \cos \theta = \cos \theta \). Recognize that this can be rearranged by factoring out \( \cos \theta \) from each term:\[ \cos \theta (\sin^2 \theta - 1) = 0. \]This factors the equation into two possible cases for solutions.

Step 2: Solve the First Case

In the first case, set the factor \( \cos \theta = 0 \). The values of \( \theta \) that satisfy \( \cos \theta = 0 \) in the interval \([0^{\circ}, 360^{\circ})\) are:- \( \theta = 90^{\circ} \)- \( \theta = 270^{\circ} \).

Step 3: Solve the Second Case

In the second case, set the second factor \( \sin^2 \theta - 1 = 0 \). Solving gives:\[ \sin^2 \theta = 1 \]\[ \sin \theta = \pm 1 \]The values of \( \theta \) that satisfy this equation in the interval \([0^{\circ}, 360^{\circ})\) are:- \( \theta = 90^{\circ} \)- \( \theta = 270^{\circ} \).

Step 4: Combine Solutions

Combine the solutions from both cases. Both cases give us \( \theta = 90^{\circ} \) and \( \theta = 270^{\circ} \). These are the complete solutions over the interval \([0^{\circ}, 360^{\circ})\).

Key Concepts

Interval Notation

Interval notation is a shorthand way of writing the set of all numbers between certain boundaries. It uses brackets to denote whether endpoints are included, and parentheses if they are not.
For example, the given problem asks for solutions over the interval \([0^\circ, 360^\circ)\). This interval includes 0 degrees but not 360 degrees.
Think of intervals as being on a number line:

• Brackets [ ]: The value at this endpoint is included. So in \([0^\circ, 360^\circ)\), 0 is a valid solution.
• Parentheses ( ): The value at this endpoint is not included. Here, 360 is not a part of the solution set.

Understanding interval notation helps clarify what values a variable, such as \(\theta\) in trigonometric problems, can take. This is crucial when evaluating trigonometric solutions over specific ranges, ensuring we only consider valid solutions within the defined boundaries.

Trigonometric Identities

Trigonometric identities are mathematical equations that relate trigonometric functions to one another. These identities are helpful to simplify expressions and solve equations.
In this problem, we deal with basic trigonometric identity knowledge:

• Pythagorean Identity: This is one of the most important identities, and it relates sine and cosine: \( \sin^2 \theta + \cos^2 \theta = 1 \).

While not directly used here, knowing this identity helps us recognize relationships in trigonometric equations. In our specific case, recognizing that \( \sin^2 \theta - 1 = 0 \) implies simplifying the equation by considering basic trigonometric values:

• For \( \sin^2 \theta = 1 \), we realize \( \sin \theta = \pm 1 \), which occurs at certain predefined angles.

Working with identities allows us to transition complex trigonometric equations into simpler forms, making them easier to solve.

Factoring Equations

Factoring is one strategic method to solve equations by rewriting them in a simpler form, which makes solutions more apparent. The process involves finding factors (expressions that multiply together) to form the original equation.
In our trigonometric problem, we started with \( \sin^2 \theta \cos \theta = \cos \theta \). The process of factoring involved:

• Recognizing \( \cos \theta \) as a common factor on both sides of the equation.
• Rewriting using this factor, giving us \( \cos \theta (\sin^2 \theta - 1) = 0 \).

When you set a product to zero like \( a \cdot b = 0 \), it means either \( a = 0 \) or \( b = 0 \). Applying this principle, we broke it into two equations:

• \( \cos \theta = 0 \)
• \( \sin^2 \theta - 1 = 0 \)

Factoring provides strategic insight into equations, transforming them into simpler equations that are easier to solve. It’s an essential skill for solving various algebraic and trigonometric problems.