91Ó°ÊÓ

A quadratic equation is generally in the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) is the unknown variable. In the context of our trigonometric equation, \( heta \) is the variable while \( ext{sin} \theta \) is the object of interest. When the equation \( 2 \sin^2 \theta - 1 = 0 \) is rearranged, it takes a similar quadratic form: \( 2 \sin^2 \theta - 1 = 0 \). By replacing \( \sin \theta \) with \( x \), the equation looks similar to the standard quadratic equation as \( 2x^2 - 1 = 0 \). Solving for \( x \), we find that \( x = \pm \frac{\sqrt{2}}{2} \). This simple transformation shows the power of understanding quadratic equations, as it lets us solve more complex expressions efficiently.

The unit circle is a fundamental concept in trigonometry. It's a circle with a radius of \( 1 \) centered at the origin of a coordinate plane. The unit circle is particularly useful because it provides a simple means of visualizing the angles and helps in understanding the trigonometric functions such as sine and cosine.

On the unit circle:

• The x-coordinate is \( \cos \theta \).
• The y-coordinate is \( \sin \theta \).

Knowing this, we can easily identify the values where the \( \sin \theta \) achieves key values like \( \pm \sqrt{2}/2 \). These points on the circle are extremely useful for solving trigonometric equations because they correspond to specific angle measures. Recognizing these standard angle positions on the unit circle allows us to quickly find solutions to equations involving trigonometric functions.

The sine function, one of the primary trigonometric functions, relates an angle in a right triangle to the ratio of the opposite side to the hypotenuse. It's periodic, meaning it repeats its values in regular intervals, specifically every \( 360^{\circ} \) or \( 2\pi \) radians.

Its range lies between \(-1\) and \(1\), hitting critical points such as:

• 0 at \(0^{\circ}\), \( 180^{\circ} \), and \( 360^{\circ} \)
• 1 at \(90^{\circ}\)
• -1 at \(270^{\circ}\)

For our exercise, recognising when \( \sin \theta = \pm \sqrt{2}/2 \) is essential, as it leads us to common angles of \(45^{\circ}\), \(135^{\circ}\), \(225^{\circ}\), and \(315^{\circ}\) on the unit circle. Understanding these relationships allows us to move seamlessly from solving the quadratic form to finding angle solutions for \( \theta \).

Finding angle solutions in the context of trigonometric equations involves determining which specific angles correspond to given sine values within a specified interval. For the problem \( 2 \sin^2 \theta - 1 = 0 \), our task is to solve \( \sin \theta = \pm \sqrt{2}/2 \) for angles \( \theta \) between \( 0^{\circ} \) and \( 360^{\circ} \).

With the knowledge of how sine values map to angles on the unit circle, we determine:

• \( \sin \theta = \frac{\sqrt{2}}{2} \) at \( \theta = 45^{\circ} \) and \( \theta = 135^{\circ} \)
• \( \sin \theta = -\frac{\sqrt{2}}{2} \) at \( \theta = 225^{\circ} \) and \( \theta = 315^{\circ} \)

These angles, typical of a rotation around the unit circle, represent a full cycle of evaluating sine values within one full rotation. By compiling these solutions, we fully solve the equation for \( \theta \) and ensure that all solutions fall within the given interval.