Problem 26 - A Graphical Approach to Precalculus with Limits

Question

Solve \((\mathbf{a}) f(x)=0,(\mathbf{b}) f(x)>0,\) and \((\mathbf{c}) f(x)<0\) over the interval \([0,2 \pi)\) $$f(x)=2 \sin x+1$$

Step-by-Step Solution

Answer

a) \(x = \frac{7\pi}{6}, \frac{11\pi}{6}\); b) \([0, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi)\); c) \((\frac{7\pi}{6}, \frac{11\pi}{6})\)."

Step 1: Solve for f(x) = 0

To find the values of \(x\) where \(f(x) = 0\), set the function equal to zero: \[ 2 \sin x + 1 = 0 \] Subtract 1 from both sides: \[ 2 \sin x = -1 \] Divide by 2: \[ \sin x = -\frac{1}{2} \] The solution for \(\sin x = -\frac{1}{2}\) within \([0, 2\pi)\) is \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\).

Step 2: Solve for f(x) > 0

We need to find where \(f(x) = 2 \sin x + 1 > 0\). Rearrange and solve the inequality: \[ 2 \sin x + 1 > 0 \] \[ 2 \sin x > -1 \] \[ \sin x > -\frac{1}{2} \] The sine function is greater than \(-\frac{1}{2}\) in the intervals \([0, \frac{7\pi}{6})\) and \( (\frac{11\pi}{6}, 2\pi)\).

Step 3: Solve for f(x) < 0

We need to determine where \(f(x) = 2 \sin x + 1 < 0\). Rearrange and solve the inequality: \[ 2 \sin x + 1 < 0 \] \[ 2 \sin x < -1 \] \[ \sin x < -\frac{1}{2} \] The sine function is less than \(-\frac{1}{2}\) in the interval \((\frac{7\pi}{6}, \frac{11\pi}{6})\).

Key Concepts

Sine Function

The sine function is one of the most fundamental trigonometric functions. It describes the y-coordinate of a point on the unit circle as the angle changes. The function is periodic, meaning it repeats its values in regular intervals. For sine, this period is \(2\pi\). It ranges from -1 to 1, reaching its maximum at 1 and minimum at -1. This inherent oscillation is due to its circular definition, as every rotation corresponds to a point on the circle.

• The sine function, \( \sin(x) \), is odd, meaning that \( \sin(-x) = -\sin(x) \).
• This function is zero at integral multiples of \(\pi\).
• The key points to remember are \(\sin(0) = 0\), \(\sin(\pi/2) = 1\), and \(\sin(3\pi/2) = -1\).

In problems involving equations like \( \sin(x) = -\frac{1}{2} \), we look for angles where the sine value equals this specific ratio. This is often solved by identifying key angles, and using symmetry properties of the function.

Interval Notation

Interval notation is a system used to denote the set of solutions within a specific range on the number line. In trigonometric problems, understanding which parts of a cycle meet the solution criteria is crucial.

• An interval such as \([a,b)\) means that \(a\) is included in the set, but \(b\) is not.
• Brackets \([\ ]\) signify inclusive bounds, while parentheses \((\ )\) are exclusive.
• This notation helps communicate the span of x-values that satisfy an equation or inequality.

For example, the solution \([0, \frac{7\pi}{6})\) shows that \(x\) values from \(0\) up to but not including \(\frac{7\pi}{6}\) satisfy the inequality \(2\sin(x) + 1 > 0\). Intervals let us succinctly express the segments of interest on the unit circle.

Inequalities

Solving inequalities in trigonometry involves finding the set of values where a trigonometric expression is either greater or less than a specific number. This often requires understanding the behavior of the functions over their cycles.

• Inequalities using sine, like \(2 \sin x + 1 > 0\), are solved by isolating \(\sin x\) and identifying intervals using the unit circle.
• The key is to determine which portions of the sine wave lie above or below the comparative value.
• If \( \sin(x) > a\), you're looking for the parts of the sine graph above the line \(y = a\).

Specific angles that relate to familiar sine values can be used to find intervals, and the use of interval notation clarifies the solutions. Knowing key trigonometric identities and values helps in determining where the inequality holds true. For example, \(\sin x > -\frac{1}{2}\) fits in portions of the cycle like \([0, \frac{7\pi}{6})\) or \((\frac{11\pi}{6}, 2\pi)\).

Unit Circle

The unit circle is a powerful tool in trigonometry, providing a geometric representation from which the values of sine and other trigonometric functions can be easily visualized and calculated. It is a circle with a radius of 1, centered at the origin of a coordinate plane.

• Angles are measured in radians, where \(2\pi\) radians completes a full circle.
• The coordinate \((\cos(\theta), \sin(\theta))\) represents a point on the circumference of the circle.
• This approach helps visualize periodic behavior and relations between angles and their trigonometric ratios.

For example, the solution to \(\sin x = -\frac{1}{2}\) involves identifying points on the unit circle where the y-coordinate is -0.5. This occurs at angles that correspond to the standard positions of \(\frac{7\pi}{6}\) and \(\frac{11\pi}{6}\). The unit circle simplifies the process of solving such equations by providing a clear visual framework.