Problem 20 - Precalculus Mathematics for Calculus

Question

An equation is given. (a) Find all solutions of the equation. (b) Find the solutions in the interval \([0,2 \pi)\) $$2 \sin 3 \theta+1=0$$

Step-by-Step Solution

Answer

Solutions are \( \frac{7\pi}{18},\frac{11\pi}{18},\frac{23\pi}{18},\frac{29\pi}{18} \) within \([0, 2\pi)\).

Step 1: Isolate the Trigonometric Function

Given the equation \(2 \sin 3\theta + 1 = 0\), move the constant term to the other side by subtracting 1 from both sides. This gives us \(2 \sin 3\theta = -1\).

Step 2: Solve for the Sine Value

Next, divide both sides by 2 to isolate \( \sin 3\theta \). This gives us \( \sin 3\theta = -\frac{1}{2} \).

Step 3: Determine General Solutions for \(\sin\theta = -\frac{1}{2}\)

The equation \( \sin u = -\frac{1}{2} \) has general solutions: \( u = \frac{7\pi}{6} + 2k\pi \) and \( u = \frac{11\pi}{6} + 2k\pi \), where \( k \) is any integer. Set \( u = 3\theta \) to reflect our problem.

Step 4: Solve for \(\theta\) from General Solutions

Substitute \( 3\theta \) back for \( u \), producing: \( 3\theta = \frac{7\pi}{6} + 2k\pi \) and \( 3\theta = \frac{11\pi}{6} + 2k\pi \). Solve these by dividing the entire equation by 3, resulting in two sets of solutions: \[ \theta = \frac{7\pi}{18} + \frac{2k\pi}{3} \] and \[ \theta = \frac{11\pi}{18} + \frac{2k\pi}{3} \].

Step 5: Identify Solutions in the Interval \([0, 2\pi)\)

For each set of solutions, substitute integers for \( k \) to find which values of \( \theta \) fall within the interval \([0, 2\pi)\). The possible values for \( k \) are checked until values exceed \( 2\pi \).Identify:- For \( \theta = \frac{7\pi}{18} + \frac{2k\pi}{3} \): - \( k = 0 \): \( \theta = \frac{7\pi}{18} \) - \( k = 1 \): \( \theta = \frac{23\pi}{18} \)- For \( \theta = \frac{11\pi}{18} + \frac{2k\pi}{3} \): - \( k = 0 \): \( \theta = \frac{11\pi}{18} \) - \( k = 1 \): \( \theta = \frac{29\pi}{18} \), which is within the interval.

Key Concepts

Sine Function

The sine function is a fundamental concept in trigonometry, representing a crucial relationship in right-angled triangles. It associates any given angle with the ratio of the length of the opposite side to the hypotenuse. In other words, for an angle \(\theta\), \(\sin \theta\) gives this ratio. The sine function is periodic, which means it repeats its values in a regular cycle. This periodic nature is important because it allows us to express solutions to equations involving sine in a repetitive pattern. The sine function's range is between \(-1\) and \(+1\), which means for any angle \(\theta\), \(-1 \leq \sin \theta \leq +1\). This feature helps in solving equations because it limits the possible solutions to within this range, allowing us to use this constraint to our advantage. A trigonometric equation like \(2 \sin 3\theta + 1 = 0\) uses transformations to make it solvable. By isolating \(\sin 3\theta\), we simplify the equation to \(\sin 3\theta = -\frac{1}{2}\), thus focusing only on the sine values that satisfy this condition.

General Solutions

In trigonometry, the idea of general solutions revolves around capturing all possible answers for an equation. Because trig functions like sine are periodic, they repeat their values over intervals, typically every \(2\pi\) for sine and cosine. This periodicity allows solutions not just once, but numerous times across different cycles.Taking our original problem, we find that \(\sin 3\theta = -\frac{1}{2}\) has solutions at specific angles on a unit circle. These angles are found using known results of the sine function, \(u = \frac{7\pi}{6}\) and \(u = \frac{11\pi}{6}\) due to their positions on the unit circle where the sine value is \(-\frac{1}{2}\).To express these as general solutions, we translate these angles to a general formula:

• \(3\theta = \frac{7\pi}{6} + 2k\pi\)
• \(3\theta = \frac{11\pi}{6} + 2k\pi\)

Here, \(k\) is any integer, representing the infinite cycles of the sine function’s periodic nature. This general formula accounts for every occurrence of the solution across all cycles. Finally, applying the division by 3 gives \(\theta\) in terms of \(k\), providing every possible solution for \(\theta\).

Trigonometric Identities

Trigonometric identities are mathematical formulas that express relationships between trigonometric functions. They are powerful tools in simplifying and solving equations. Identities like Pythagorean identities, angle sum identities, and double-angle identities help us tackle complex trigonometric equations efficiently.In the given exercise, while explicit identities were not directly used, the solving process inherently relies on the properties and known behaviors of the sine function. For example, knowing that \(\sin \theta\) reaches \(-\frac{1}{2}\) at specific angles like \(\frac{7\pi}{6}\) and \(\frac{11\pi}{6}\) helps find solutions using the identities:

• \(\sin(\pi + \theta) = -\sin \theta\)
• \(\sin(2\pi + \theta) = \sin \theta\)

These identities confirm the period and the sign change in sine at these angles. They underpin our understanding of where these solutions exist and ensure mathematical consistency in cyclic solutions. Trigonometric identities further assure that solutions are not just correct, but are comprehensive within their intended domain, producing valid angles that fulfill the original equation across its full range of operation.