Problem 16 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

\text { Find all } t \in[-\pi, \pi] \text { such that } 2 \sin ^{2} t-3 \sin t+1=0 \text { . }

Step-by-Step Solution

Answer

The solutions for the equation are \(t = \frac {\pi}{6}\), \(t = \frac {5\pi}{6}\), and \(t = \frac {\pi}{2}\).

Step 1: Recognize the Structure

Observe that the equation \(2 \sin ^{2} t - 3 \sin t + 1 = 0\) follows a quadratic structure \(a \sin^2 t + b \sin t + c = 0\), in which a is 2, b is -3, and c is 1.

Step 2: Apply the Quadratic Formula

Solving the quadratic equation for the variable \(\sin t\) can be done using the quadratic formula \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). So, \(\sin t = \frac {3 \pm \sqrt{{(-3)}^2 - 4*2*1}}{2*2} = \frac {3 \pm \sqrt{1}}{4}\). This leaves us with two possible solutions for \(\sin t\), namely \(\frac {1}{2}\) and \(1\).

Step 3: Interpret Results as Angle Measurements

The values \(\sin t = \frac {1}{2}\) and \(\sin t = 1\) correspond to different angles in the unit circle. For \(\frac {1}{2}\), they are \(t = \frac {\pi}{6}\) and \(t = \frac {5\pi}{6}\). While for \(1\), it is \(t = \frac {\pi}{2}\). All angles are measured in radians.

Step 4: Checking the Results

All the solutions \(t = \frac {\pi}{6}\), \(t = \frac {5\pi}{6}\), and \(t = \frac {\pi}{2}\) are within the given interval \([- \pi, \pi]\). So, these are the solutions for \(t\) of the given equation.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving equations of the form \( ax^2 + bx + c = 0 \). It's especially useful when equations do not factor easily. The formula is expressed as \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

This formula helps find the roots of the polynomial, which are the values of \( x \) that make the equation true. In trigonometric problems, like the one above, the quadratic formula is applied to \( \sin(t) \). Here, \( a = 2 \), \( b = -3 \), and \( c = 1 \), leading up to \( \sin(t) = \frac{1}{2} \) and \( \sin(t) = 1 \).

For trigonometric equations, these solutions aren't complete until we interpret them with respect to angles. That leads us seamlessly into our next concept.

Sine Function

The sine function is vital in solving trigonometric equations because it relates angles to ratios in a right triangle. For any angle \( t \) on the unit circle,

• \( \sin(t) = \frac{\text{opposite side}}{\text{hypotenuse}} \)

However, on the unit circle, the radius is 1, simplifying \( \sin(t) \) to the y-coordinate of the point on the unit circle.

In our exercise, we calculated \( \sin(t) = \frac{1}{2} \) and \( \sin(t) = 1 \). We then use these values to determine the angles \( t \) that satisfy this.

For \( \sin(t) = \frac{1}{2} \), the possible angles are \( t = \frac{\pi}{6} \) and \( t = \frac{5\pi}{6} \). When \( \sin(t) = 1 \), the angle is \( t = \frac{\pi}{2} \). These values fall within \([-\pi, \pi]\), completing the interpretation of this function for our purposes.

Unit Circle

The unit circle is a central concept in trigonometry used to define sine and other trigonometric functions. It has a radius of 1 and is centered at the origin of a coordinate plane.

Every point on the unit circle represents an angle from the x-axis and can be expressed as \((\cos(t), \sin(t))\). The complete circle goes from \(-\pi\) to \(\pi\), covering a full rotation.

In the context of our problem, we find angles where \( \sin(t) \) matches our solutions:

• \( \sin(t) = \frac{1}{2} \) at \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \)
• \( \sin(t) = 1 \) at \( \frac{\pi}{2} \)

These represent positions on the unit circle where the y-coordinates are \( \frac{1}{2} \) and 1, respectively.

The unit circle is invaluable for visualizing how these angles relate back to sine values, ensuring that we correctly identify all possible solutions to the equation.