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A quadratic equation is a second-degree polynomial equation in the form of \( ax^2 + bx + c = 0 \). Understanding how to rearrange and solve such equations is essential in mathematics, especially when applied to physical problems like trigonometry.
In our given problem, the equation \( 2 \sin^2 \theta - 3 \sin \theta = -1 \) is rearranged to the standard quadratic form \( 2 \sin^2 \theta - 3 \sin \theta + 1 = 0 \). This involves simple algebraic manipulation, such as adding or subtracting terms on both sides of the equation. In this case, adding "+1" to both sides transformed the equation into a standard quadratic format.

• Standard form makes it easier to apply the quadratic formula.
• The coefficients here are important: identify \( a = 2 \), \( b = -3 \), and \( c = 1 \).

By substituting \( \sin \theta \) with another variable \( x \), it simplifies the process, letting us employ the quadratic formula easily.

The sine function, commonly denoted as \( \sin \theta \), is a fundamental trigonometric function. It relates an angle in a right triangle to the ratio of the opposite side over the hypotenuse. The function is periodic, repeating every \( 360^\circ \) or \( 2\pi \) radians.
The key properties of the sine function include:

• Its range is between \(-1\) and \(1\).
• A full cycle within \(0^\circ\) to \(360^\circ\) goes from \(0\) to \(1\) and back to \(0\), continuing to \(-1\) and returning to \(0\).

In the trigonometric equation context, \( \sin \theta \) values determine potential angle solutions. For instance, when solving \( \sin \theta = x \), where \( x \) represents our quadratically derived solutions, you seek angles that produce those specific sine values. In our problem, two values, \( x = 1 \) and \( x = 0.5 \), lead to identifiable angle solutions.

Finding angle solutions involves determining which angles, within a specified range, satisfy the equation. Given the constraint \(0^\circ \leq \theta < 360^\circ\), angle solutions leverage the properties and periodicity of trigonometric functions.
With \( \sin \theta = 1 \), the angle solution is unique: \( \theta = 90^\circ \), as it is the only angle within one period where the sine function achieves its maximum value.
For \( \sin \theta = 0.5 \), the potential angles involve evaluating within the unit circle, resulting in two solutions: \( 30^\circ \) and \( 150^\circ \). These correspond to the sine values being positive in the first and second quadrants.

• Consider all quadrants: Sine is positive in both the first and second quadrants.
• Each solution falls within \(0^\circ\) to \(360^\circ\), ensuring they are valid within our given range.

Piecing together these solutions results in \( \theta = 30^\circ, 90^\circ, 150^\circ \), all satisfying the trigonometric and quadratic conditions.