Problem 32 - Precalculus

Question

For the following exercises, let \(f(x)=\sin x\) On \([0,2 \pi),\) solve \(f(x)=\frac{1}{2}\)

Step-by-Step Solution

Answer

The solutions are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).

Step 1: Identify the Equation

The given function is \( f(x) = \sin x \) and we need to solve \( f(x) = \frac{1}{2} \). Thus, we are looking for \( x \) such that \( \sin x = \frac{1}{2} \).

Step 2: Determine General Solutions

We know from the unit circle that \( \sin x = \frac{1}{2} \) at angles \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \) in the range \( 0 \leq x < 2\pi \).

Step 3: Check within the Interval

Both solutions \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \) are within the interval \([0, 2\pi)\). There are no additional solutions in this interval for \( \sin x = \frac{1}{2} \) as only the first and second quadrants are involved.

Key Concepts

Unit Circle

The unit circle is a fundamental concept in trigonometry, crucial for understanding trigonometric equations. It is a circle with a radius of 1 centered at the origin of a coordinate plane. On the unit circle, each point represents

• an angle measured in radians from the positive x-axis,
• the x-coordinate as the cosine of the angle,
• the y-coordinate as the sine of the angle.

To solve trigonometric equations like the one given, it’s important to understand the significance of specific angles where certain sine and cosine values occur. For instance, the sine function equals 0.5 specifically at

• the angles \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \) on the unit circle.

The unit circle provides an easy reference to these angles, making calculations simpler when solving similar trigonometric problems.

Sine Function

The sine function is one of the primary trigonometric functions and is denoted by \( \sin x \). For an angle \( x \), the sine function is defined as the y-coordinate of the point on the unit circle. The output of the sine function ranges from -1 to 1. This means:

• When \( \sin x = 1 \), the angle is \( \frac{\pi}{2} \),
• When \( \sin x = -1 \), the angle is \( \frac{3\pi}{2} \).

For solving the equation \( \sin x = \frac{1}{2} \), we look for angles where the sine value equals 0.5. As noted earlier, the angles are \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \). Here are some properties of the sine function that help in solving such equations:

• It repeats every \( 2\pi \), known as the periodicity of the sine function.
• The sine function is positive in the first and second quadrants of the unit circle.
• Knowing these properties helps predict the behavior of the sine function over its domain.

The knowledge of these points allows us to solve equations involving the sine function within specified intervals.

Interval Notation

Interval notation is a way of representing a range of numbers, often a segment of the real number line. In trigonometric problems, it helps clearly define the specific part of the domain we're interested in examining. For example,

• \([0, 2\pi)\) indicates all numbers starting at 0 up to, but not including, \(2\pi\).
• The square bracket \([\) means the boundary value is included, while the parenthesis \()\) means it is not.

In this exercise, considering the interval

• \([0, 2\pi)\) narrows down and specifies where solutions to \( \sin x = \frac{1}{2} \) should be found.
• Because sine function values repeat after each complete circle (\(2\pi\)), interval notation simplifies expressing the set of angles where the function meets a particular condition.

By understanding interval notation, we can succinctly express conditions like the one given in the exercise.