Problem 33 - 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. $$2 \tan ^{2} \theta \sin \theta-\tan ^{2} \theta=0$$

Step-by-Step Solution

Answer

The solutions are \(0^\circ, 180^\circ, 30^\circ,\) and \(150^\circ\).

Step 1: Factor the Equation

First, notice that both terms in the equation have a common factor of \( \tan^2 \theta \). Factor the equation: \[ \tan^2 \theta (2 \sin \theta - 1) = 0 \].

Step 2: Solve Each Factor Separately

Set each factor equal to zero to find the solutions. This gives us two equations: \( \tan^2 \theta = 0 \) and \( 2 \sin \theta - 1 = 0 \).

Step 3: Solve \( \tan^2 \theta = 0 \)

Since \( \tan^2 \theta = 0 \) when \( \tan \theta = 0 \), solve for \( \theta \). Recall that \( \tan \theta = 0 \) at \( \theta = 0^\circ \) and \( \theta = 180^\circ \) within \( [0^\circ, 360^\circ) \).

Step 4: Solve \( 2 \sin \theta - 1 = 0 \)

Rearrange to find \( \sin \theta = \frac{1}{2} \). The angles where \( \sin \theta = \frac{1}{2} \) within \( [0^\circ, 360^\circ) \) are \( \theta = 30^\circ \) and \( \theta = 150^\circ \).

Step 5: Combine All Solutions

From both factored terms, we collect all solutions obtained: \( \theta = 0^\circ, 180^\circ, 30^\circ, \) and \( 150^\circ \).

Key Concepts

Factoring Equations

Factoring equations is a helpful strategy when dealing with trigonometric expressions. It simplifies the problem by reducing it to simpler parts that can be solved separately. In the exercise provided, both terms in the equation share a common factor, \(\tan^2 \theta\). By factoring out this common term, the original equation \(2 \tan ^{2} \theta \sin \theta-\tan ^{2} \theta=0\) becomes \(\tan^2 \theta (2 \sin \theta - 1) = 0\).

Factoring can be seen as breaking down the expression into "pieces" that multiply to give the original problem. The idea is that if a product of terms equals zero, then at least one of the terms must be zero. This approach allows us to split the problem into manageable parts: \(\tan^2 \theta = 0\) and \(2 \sin \theta - 1 = 0\). Solving each of these simpler equations separately can give all possible solutions for the original equation. This method is analogous to solving quadratic equations through factorization in algebra.

• Look for common factors to simplify complex expressions.
• Becomes easier to solve with each term set to zero separately.
• Helps in breaking down complex trigonometric functions.

Tangent Function

The tangent function, denoted as \(\tan \theta\), is one of the fundamental trigonometric functions. Its role in solving trigonometric equations stems from its oscillating nature over periodic intervals. In this exercise, the equation \(\tan^2 \theta = 0\) was derived after factoring.

Recognizing when \(\tan \theta = 0\) is crucial: tangent is zero when the angle \(\theta\) leads to a sine of zero, and cosine is non-zero. This happens at specific points: \(\theta = 0^{\circ}\) and \(180^{\circ}\) within the interval \([0^{\circ}, 360^{\circ})\).

• \(\tan \theta\) is zero at integer multiples of \(180^{\circ}\).
• Keeps repeating every \(180^{\circ}\).
• Useful in identifying angle solutions when factored into equations.

Sine Function

The sine function, represented as \(\sin \theta\), is essential in trigonometry due to its role in representing periodic oscillations. In this exercise, after setting the factor equation \(2 \sin \theta - 1 = 0\), solving for \(\sin \theta\) gives us the value \(\frac{1}{2}\).

Identifying \(\theta\) values that satisfy \(\sin \theta = \frac{1}{2}\) involves recognizing familiar angles from trigonometric circles. Within the interval \([0^{\circ}, 360^{\circ})\), these angles are \(30^{\circ}\) and \(150^{\circ}\). These can be remembered through common trigonometric ratios or by using the unit circle. The sine function indicates the height of a point on the unit circle, which corresponds to these angles.

• \(\sin \theta = \frac{1}{2}\) at \(30^{\circ}\) and \(150^{\circ}\).
• Represents vertical projection in unit circle terms.
• Key in solving trigonometric identities and equations.

Solution of Equations

Solving trigonometric equations involves finding specific angle values that make the equation true. The process typically includes simplifying by factoring, if possible, and solving each factor separately. Each step should adhere to the domain or interval of the problem statement. In this case, giving answers within \([0^{\circ}, 360^{\circ})\) ensures completeness within a full rotation cycle.

After factoring the equation and handling \(\tan^2 \theta = 0\) and \(2 \sin \theta - 1 = 0\), solutions collect integers and specific angle solutions: \(\theta = 0^{\circ}, 180^{\circ}, 30^{\circ}, 150^{\circ}\).

Each of these points represents a valid solution that can be checked directly into the original equation to verify their validity.

• Consider intervals to confirm solution appropriateness.
• Solve factored parts and combine solutions for completeness.
• Ensure that solutions reflect the periodic nature of trigonometric functions.