Problem 11 - Calculus with Concepts in Calculus

Question

Solve the inequality for \(x\) in \([0,2 \pi)\). $$ \sin x>-\frac{1}{2} $$

Step-by-Step Solution

Answer

The solutions are \(x \in [0, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi)\)."

Step 1: Understand the given inequality

We need to solve the inequality \( \sin x > -\frac{1}{2} \) for \( x \) in the interval \( [0, 2\pi) \). This means we need to find all the values of \( x \) where the sine of \( x \) is greater than \(-\frac{1}{2}\).

Step 2: Identify where \( \sin x = -\frac{1}{2} \)

The sine of an angle equals \(-\frac{1}{2}\) at specific angles within each period. In the unit circle, these angles for \( \sin x = -\frac{1}{2} \) are given by \( x = \frac{7\pi}{6} \) and \( x = \frac{11\pi}{6} \).

Step 3: Determine intervals of inequality

The sine function is continuous and periodic with period \(2\pi\). Between the points where \(\sin x = -\frac{1}{2}\), i.e., \(x = \frac{7\pi}{6} \) and \( x = \frac{11\pi}{6} \), \(\sin x\) is naturally greater than \(-\frac{1}{2}\). This results in two segments within the interval \([0, 2\pi)\): \(x \in [0, \frac{7\pi}{6})\) and \(x \in (\frac{11\pi}{6}, 2\pi)\).

Step 4: Combine the solution intervals

Combining the solutions from the previous step, the values of \(x\) which satisfy the inequality \(\sin x > -\frac{1}{2}\) are \([0, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi)\).

Key Concepts

unit circle

The unit circle is a fundamental tool in trigonometry that helps us understand angle measurements and their corresponding trigonometric functions like sine and cosine. It is a circle with a radius of one, centered at the origin of a coordinate plane. By keeping the radius as one unit, every point on the unit circle has coordinates represented by

• For any angle \( \theta \) measured from the positive x-axis, the x-coordinate is \( \cos \theta \)
• The y-coordinate is \( \sin \theta \).

Understanding the unit circle is crucial when solving trigonometric inequalities as it allows us to visually identify and calculate the sine or cosine of specific angles. In our exercise, we're interested in the points where the sine of the angle is more than \(-\frac{1}{2}\).
To do this, we locate the angles on the unit circle that correspond to the points where \( \sin x = -\frac{1}{2} \), which are at \( x = \frac{7\pi}{6} \) and \( x = \frac{11\pi}{6} \).
By analyzing the continuous nature of the sine function on the unit circle, we can see the behavior above these reference points, ensuring we find all segments where the inequality holds.

sine function

The sine function is one of the foundational elements in trigonometry, often symbolized as \( \sin(x) \). It describes a wave that oscillates between -1 and 1, repeating every \( 2\pi \) radians (one full circle on the unit circle). This periodic feature of the sine function allows us to predict its values for any angle.

• In the first half of the function’s cycle, from \( 0 \) to \( \pi \), the sine values increase from 0 to 1 and then decrease back to 0.
• In the second half, from \( \pi \) to \( 2\pi \), the sine values go from 0 to -1 and back to 0.

When examining the inequality \( \sin x > -\frac{1}{2} \), we are looking for the ranges where the y-coordinate of the unit circle (representing \( \sin x \)) exceeds \(-\frac{1}{2}\).
The values of \( x \) where \( \sin x = -\frac{1}{2} \) act as boundary points, marking transitions in the sine function's graph. By understanding the continuous wave-like structure of the sine graph, we predict where the function remains above our boundary between \( [0, 2\pi) \).

interval notation

Interval notation is a way of denoting intervals on the real number line, which is particularly useful for expressing ranges of solutions in mathematics, like in inequalities. This notation helps us indicate which portions of a range are included or excluded by using different brackets.

• A square bracket \([ \text{ or } ]\) means that the endpoint is included in the interval, known as 'closed'.
• A parenthesis \(( \text{ or } )\) indicates an endpoint is excluded from the interval, referred to as 'open'.

In the situation where we solved \( \sin x > -\frac{1}{2} \), we used interval notation to find and represent the parts of the interval \([0, 2\pi)\) where the sine function holds above the threshold.
The solution is expressed as \([0, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi)\).
This tells us that every point from \(0\) to \(\frac{7\pi}{6}\) is valid (including \(0\), but not \(\frac{7\pi}{6}\)), combined with values from just above \(\frac{11\pi}{6}\) up to but not including \(2\pi\).
Understanding how to use interval notation allows us to quickly and clearly convey the domain of solutions we are working with.