Problem 29 - Precalculus with Calculus Previews

Question

Find all solutions of the given trigonometric equation if \(x\) is a real number and \(\theta\) is an angle measured in degrees. $$ 2 \sin 3 \theta=1 $$

Step-by-Step Solution

Answer

The solutions are \( \theta = 10^\circ + 120^\circ k \) and \( \theta = 50^\circ + 120^\circ k \), where \( k \) is an integer.

Step 1: Isolate the Sine Function

The original equation is given by \[ 2 \sin 3\theta = 1 \]. To isolate \( \sin 3\theta \), divide both sides by 2:\[ \sin 3\theta = \frac{1}{2} \].

Step 2: Solve for the Angle

We know from trigonometry that \( \sin \theta = \frac{1}{2} \) when \( \theta = 30^\circ + k \cdot 360^\circ \) or \( \theta = 150^\circ + k \cdot 360^\circ \), where \( k \) is an integer. Thus, set \( 3\theta = 30^\circ + 360^\circ k \) and \( 3\theta = 150^\circ + 360^\circ k \).

Step 3: Solve for \( \theta \) in General Form

Divide each general solution by 3: - For \( 3\theta = 30^\circ + 360^\circ k \), we have: \[ \theta = 10^\circ + 120^\circ k \]. - For \( 3\theta = 150^\circ + 360^\circ k \), we have: \[ \theta = 50^\circ + 120^\circ k \].

Step 4: Compile the General Solutions

Combine these results to find the general solutions for \( \theta \):\[ \theta = 10^\circ + 120^\circ k \quad \text{and} \quad \theta = 50^\circ + 120^\circ k \].Here, \( k \) is any integer.

Key Concepts

Sine Function

The sine function is one of the primary trigonometric functions, alongside cosine and tangent. It is essential for understanding various aspects of trigonometry, especially in relation to angles and periodic functions. When you see the sine function, it often pertains to a right-angled triangle, relating to the ratio of the length of the side opposite the angle to the hypotenuse (the longest side).In our exercise, the sine function is expressed as \( \sin 3\theta \). The equation \( 2 \sin 3\theta = 1 \) demonstrates how sine functions can be embedded within equations, leading to solutions that aren't purely numerical, but rather angle-based. When isolating the sine term, you rearrange to find \( \sin 3\theta = \frac{1}{2} \). This preparation is key before delving into solving the equation towards specific angles or general solutions.

Angle Measurement

Angles are measured in degrees (°) or radians, and trigonometric functions often work within these measurements. Degree measures, which are used in the problem, offer a straightforward understanding because they segment a circle into 360 equal parts for a full rotation.In this problem, angle measurement is crucial when identifying values of \( \theta \). Particularly, knowing what angles yield certain sine values helps pinpoint solutions. For instance, \( \sin \theta = \frac{1}{2} \) at angles like \( 30° \) and \( 150° \), which are derived from the unit circle or trigonometric identities. Recognizing this helps set up the equations \( 3\theta = 30° + 360°k \) and \( 3\theta = 150° + 360°k \), allowing you to further solve for \( \theta \).

General Solutions

Understanding general solutions for trigonometric equations is essential for determining not just a single solution, but all possible solutions. This method allows for the identification of all angles that satisfy the equation within the constraints of sine, cosine, or other trigonometric functions.In our exercise, the general solution involves dividing the angle expressions by 3, leading to \( \theta = 10° + 120°k \) and \( \theta = 50° + 120°k \). Here, \( k \) represents any integer, effectively capturing the infinite nature of cycles or rotations seen in circular functions. This general solution forms because the sine function is periodic, repeating its values over regular intervals (\(360°\) in this case) and hence why solutions recur every \( 120° \). By providing a formula that incorporates \( k \), we capture all such coterminal angles, addressing every possible case efficiently.