Problem 4 - Algebra 2

Question

Find all solutions of each equation for the given interval. \(3 \sin ^{2} \theta-\cos ^{2} \theta=0 ; 0 \leq \theta<\frac{\pi}{2}\)

Step-by-Step Solution

Answer

\( \theta = \frac{\pi}{6} \) is the solution.

Step 1: Use Trigonometric Identity

We can use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to express \( \cos^2 \theta \) in terms of \( \sin^2 \theta \). Substitute \( \cos^2 \theta = 1 - \sin^2 \theta \) into the equation: \( 3\sin^2 \theta - (1 - \sin^2 \theta) = 0 \).

Step 2: Simplify the Equation

Simplify the equation: \( 3\sin^2 \theta - 1 + \sin^2 \theta = 0 \). Combine like terms to get \( 4\sin^2 \theta = 1 \).

Step 3: Solve for \( \sin^2 \theta \)

Divide both sides by 4: \( \sin^2 \theta = \frac{1}{4} \). Take the square root of both sides to find \( \sin \theta = \pm \frac{1}{2} \).

Step 4: Determine \( \theta \) that satisfies \( \sin \theta = \frac{1}{2} \)

Within the interval \( 0 \leq \theta < \frac{\pi}{2} \), find \( \theta \) such that \( \sin \theta = \frac{1}{2} \). The angle that satisfies this is \( \theta = \frac{\pi}{6} \).

Step 5: Exclude \( \sin \theta = -\frac{1}{2} \)

In the interval \( 0 \leq \theta < \frac{\pi}{2} \), \( \sin \theta \) cannot be negative, so we discard solutions for \( \sin \theta = -\frac{1}{2} \).

Key Concepts

Trigonometric Identity

Trigonometric identities are equations that involve trigonometric functions and are true for any angle. They are a powerful tool in solving trigonometric equations. One such identity is the Pythagorean identity:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

This identity allows us to express one trigonometric function in terms of another. In the given exercise, we use this identity to express \( \cos^2 \theta \) in terms of \( \sin^2 \theta \). By rearranging the identity, we get:\( \cos^2 \theta = 1 - \sin^2 \theta\)This substitution simplifies the equation and makes it easier to solve. Understanding these identities is key to effectively addressing trigonometric equations.

Interval Solutions

When solving trigonometric equations, it is important to consider the interval in which solutions must be found. In this exercise, the interval is specified as

• \(0 \leq \theta < \frac{\pi}{2}\)

This means we are looking for angles \(\theta\) in the first quadrant, where both sine and cosine are positive. The interval dictates which potential solutions are valid.
For example, even though \(\sin \theta = \pm \frac{1}{2} \) are solutions algebraically, angles corresponding to negative sine are excluded due to the interval restriction. Therefore, it's crucial to carefully evaluate the solutions within the given range, considering the signs and the angles involved.

Simplification Techniques

Simplification techniques help reduce complex trigonometric equations into more manageable forms. In this exercise, simplifying involves several key steps:

• Substituting the trigonometric identity \(\cos^2 \theta = 1 - \sin^2 \theta\).
• Combining like terms: \(3\sin^2 \theta + \sin^2 \theta - 1 = 0\) simplifies to \(4\sin^2 \theta = 1\).
• Solving for \(\sin^2 \theta\): dividing both sides by 4 gives \(\sin^2 \theta = \frac{1}{4}\).

Finally, taking the square root to find \(\sin \theta\) yields \(\sin \theta = \pm \frac{1}{2}\). Each step of simplification helps isolate the variable \(\theta\) and makes it easier to find solutions in the specified interval. Use these techniques to break down steps and walk toward the solution logically.

Sine Function

The sine function is one of the fundamental trigonometric functions. It maps angles to a range of values between -1 and 1. In the context of the current exercise, we are concerned with the solutions of the equation \( \sin \theta = \pm \frac{1}{2} \).
Focusing on the positive case, \(\sin \theta = \frac{1}{2}\), within the interval \(0 \leq \theta < \frac{\pi}{2}\), the solution corresponds to

• \(\theta = \frac{\pi}{6}\)

This is because \(\frac{\pi}{6} = 30^\circ\), which is a well-known special angle where the sine value is \(\frac{1}{2}\).
Understanding the behavior of the sine function and common values associated with special angles helps solve equations quickly. These angles are frequently used in various geometry and analytical problems, making them valuable to memorize.