Problem 12 - Precalculus: Functions and Graphs

Question

Find all solutions of the equation. $$2 \sin 3 \theta+\sqrt{2}=0$$

Step-by-Step Solution

Answer

The solutions are \( \theta = \frac{5\pi}{12} + \frac{2k\pi}{3} \) and \( \theta = \frac{7\pi}{12} + \frac{2k\pi}{3} \).

Step 1: Isolate the Sine Function

Start with the given equation: \( 2 \sin 3\theta + \sqrt{2} = 0 \) Subtract \( \sqrt{2} \) from both sides to isolate the sine function:\( 2 \sin 3\theta = -\sqrt{2} \)

Step 2: Solve for Sine

Divide each side of the equation by 2 to solve for sine:\( \sin 3\theta = -\frac{\sqrt{2}}{2} \)

Step 3: Determine Angles Where the Sine Function is Negative

The sine function equals \(-\frac{\sqrt{2}}{2}\) at angles \(\frac{5\pi}{4}\) and \(\frac{7\pi}{4}\) in the unit circle. This corresponds to the reference angle \(\frac{\pi}{4}\), but located in the third and fourth quadrants.

Step 4: General Solution for Angles

Therefore, the general solutions for \(3\theta\) are given by:\( 3\theta = \frac{5\pi}{4} + 2k\pi \) and \( 3\theta = \frac{7\pi}{4} + 2k\pi \),where \( k \) is any integer.

Step 5: Solve for \( \theta \)

Divide each part of the general solution by 3 to solve for \(\theta\):\( \theta = \frac{5\pi}{12} + \frac{2k\pi}{3} \) and \( \theta = \frac{7\pi}{12} + \frac{2k\pi}{3} \) for integer \( k \).

Key Concepts

Sine Function

The sine function is a vital element of trigonometry. It originates from the unit circle concept and helps in finding the relationship between an angle and a right-angled triangle's sides.
It's represented mathematically as \( \sin\theta \), where \( \theta \) is an angle. The sine of an angle is the y-coordinate of a point on the unit circle.

• The sine function is periodic with a period of \( 2\pi \).
• Its value ranges between -1 and 1.
• It is positive in the first and second quadrants and negative in the third and fourth quadrants.

In solving equations like \( \sin 3\theta = -\frac{\sqrt{2}}{2} \), understanding the sine function's periodicity and range is crucial. This helps determine when and where the sine function takes specific values.

Unit Circle

The unit circle is a circle with a radius of 1 centered at the origin of a coordinate system. It plays a significant role in trigonometry, as it helps us visualize how the sine, cosine, and tangent functions work.
The unit circle relates angle measures (in radians) with coordinates on the circle.

• The x-coordinate represents the cosine of the angle.
• The y-coordinate represents the sine of the angle.

When solving \( \sin 3\theta = -\frac{\sqrt{2}}{2} \), the unit circle helps us identify the angles corresponding to that sine value. Specifically,
this occurs when the angle's reference angle is \( \frac{\pi}{4} \) but lies in the third and fourth quadrants, thereby taking the values of \( \frac{5\pi}{4} \) and \( \frac{7\pi}{4} \), respectively.

Reference Angle

The reference angle is the acute angle that a given angle makes with the x-axis. It assists in understanding which quadrant an angle lies within the unit circle.
Reference angles are essential when working with trigonometric functions since they provide insight into how trigonometric values behave in different quadrants.

The reference angle for \( \sin 3\theta = -\frac{\sqrt{2}}{2} \) is \( \frac{\pi}{4} \). This is due to the symmetry of the sine function in the unit circle.
In the unit circle:

• In the first quadrant, the angle equals the reference angle.
• In the second quadrant, the angle is \( \pi - \text{Reference Angle} \).
• In the third quadrant, the angle is \( \pi + \text{Reference Angle} \).
• In the fourth quadrant, the angle is \( 2\pi - \text{Reference Angle} \).

Understanding these relationships helps determine the general solutions for trigonometric equations.

General Solution

The general solution of a trigonometric equation refers to all possible solutions that satisfy the equation. Knowing the general solution is crucial for finding every angle that fits within the scope of an equation.
For our problem, the general solution for \( 3\theta \) when \( \sin 3\theta = -\frac{\sqrt{2}}{2} \) utilizes the knowledge of where the sine function is negative on the unit circle.

• For the third quadrant, the equation is \( 3\theta = \frac{5\pi}{4} + 2k\pi \).
• For the fourth quadrant, the equation is \( 3\theta = \frac{7\pi}{4} + 2k\pi \).

Here, \( k \) is any integer, representing the periodicity of the sine function. To find \( \theta \), divide these equations by 3. Therefore, the general solutions are:
\( \theta = \frac{5\pi}{12} + \frac{2k\pi}{3} \) and \( \theta = \frac{7\pi}{12} + \frac{2k\pi}{3} \). This covers all angles meeting the equation's criteria.