Problem 40 - Precalculus with Limits

Question

In Exercises 39-44, solve the multiple-angle equation. \( \sin 2x = -\dfrac{\sqrt{3}}{2} \)

Step-by-Step Solution

Answer

The solution to the equation \(\sin 2x = -\dfrac{\sqrt{3}}{2}\) in the interval [0, 2pi] is \(x = 2\pi/3, 5\pi/6\).

Step 1: Identify the reference angle

We start by identifying the reference angle. The reference angle \(\theta = \frac{\pi}{3}\), \(\frac{4\pi}{3}\) has a sine of \(\frac{\sqrt{3}}{2}\). From the equation, we know that sine is negative. In the unit circle, sine is negative in the 3rd and 4th quadrants. Therefore, the angle we are looking for could be in either of these quadrants.

Step 2: Solve for solutions in the interval [0, 2pi]

We can solve for solutions in the interval [0, 2pi] equal to the values of \(2x\). In the third quadrant, the corresponding angle to our reference angle is \(4\pi/3\), and in the fourth quadrant, the corresponding angle is \(5\pi/3\). So the solutions for \(2x\) in the range from [0, 2pi] are \(4\pi/3\) and \(5\pi/3\).

Step 3: Solve for 'x'

Finally, we divide the two solutions of \(2x\) by 2 to solve for 'x'. For \(2x = 4\pi/3\), dividing by 2 yields \(x = 2\pi/3\), and for \(2x = 5\pi/3\), dividing by 2 yields \(x = 5\pi/6\). Therefore, the solutions for the given equation are \(x = 2\pi/3, 5\pi/6\) in the interval [0, 2pi].

Key Concepts

Trigonometric Functions

Trigonometric functions help us describe the relationships between the angles and sides of triangles, especially right-angled ones. In our exercise, we focus on the sine function, denoted as \( \sin \theta \).

• The sine of an angle in a right triangle gives us the ratio of the length of the opposite side to the hypotenuse.
• Sine has properties characterized by its periodicity, repeating every \( 2\pi \) radians or 360 degrees.
• It can take values between -1 and 1, inclusive.

In the given equation, \( \sin 2x = -\dfrac{\sqrt{3}}{2} \), we're dealing with a multiple-angle sine equation. The angle \( 2x \) is a multiple of a smaller angle, \( x \), indicating it's been "stretched out." Therefore, we must find where the sine of these larger angles reaches the specified value of \( -\dfrac{\sqrt{3}}{2} \). Understanding trigonometric functions is crucial as it enables us to solve for unknown angles using known sine values. This is often achieved by employing a reference angle and considering the behavior of the sine function across different quadrants of the unit circle.

Unit Circle

The unit circle is a powerful tool in trigonometry for understanding angles and trigonometric functions. It is a circle with a radius of 1 centered at the origin of a coordinate plane.

• Angles in the unit circle are measured in radians, where a full circle is \( 2\pi \) radians, equivalent to 360 degrees.
• The coordinates of any point on the unit circle \((\cos \theta, \sin \theta)\) provide the cosine and sine of the angle \( \theta \).
• Sine and cosine values correspond to the y-coordinate and x-coordinate of the point on the unit circle, respectively.

In our specific case, we need to find where the sine of angle \( 2x \) equals \(-\dfrac{\sqrt{3}}{2}\).
To do this, we utilize the nature of the unit circle and identify that sine is negative in the third and fourth quadrants. Angles in these quadrants complement the idea of reference angles and help achieve the desired sine value. The unit circle allows us to visually recognize how angles relate in a full rotation and simplifies the solution process by pointing us to exact positions where the required sine values hold true.

Reference Angle

A reference angle helps us solve trigonometric equations efficiently, especially when dealing with periodic functions like sine. A reference angle is usually the smaller or "primary" angle from which other related angles are derived.

• It is always a positive acute angle (≤ 90 degrees or ≤ \( \pi/2 \) radians).
• For sine, the reference angle provides information on which larger angles can have specified sine values when considering their position within different quadrants on the unit circle.

In the exercise, we know that \( \sin \theta = \dfrac{\sqrt{3}}{2} \) occurs commonly at angles such as \( \pi/3 \) or \( 60^{\circ} \).
When considering where sine takes the value of \(-\dfrac{\sqrt{3}}{2}\), we identify it happens in the third and fourth quadrants due to its negative nature, correlating to reference angles like \( 4\pi/3 \) and \( 5\pi/3 \). Recognizing these reference angles allows us to find solutions to multi-angle equations like \( \sin 2x = -\dfrac{\sqrt{3}}{2} \) with ease. By knowing where sine is positive or negative in the unit circle, we can derive the required solutions in any quadrant.