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In trigonometry, the Pythagorean identity is a fundamental equation that relates three key trigonometric functions: sine, cosine, and tangent. It states that for any angle \( x \), the square of the sine of the angle plus the square of the cosine of the angle equals one. Mathematically, this is written as:\[\sin^2 x + \cos^2 x = 1\]This identity is immensely useful for solving trigonometric equations because it allows you to express one function in terms of another.

• For example, in the given problem, the identity was used to replace \( \cos^2 x \) with \( 1 - \sin^2 x \).
• This substitution simplifies the equation, enabling further algebraic manipulations.

By utilizing the Pythagorean identity, you transform complex trigonometric equations into easier algebraic forms, a vital step in trigonometry problem-solving.

Quadratic equations are polynomial equations of degree two, typically expressed in the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Solving quadratic equations can be done using various methods, including factoring, completing the square, or the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In the original solution, the trigonometric equation was rewritten as a quadratic equation in terms of \( \sin x \). This turns the trigonometric problem into a purely algebraic one:

• First, rewrite the equation using the Pythagorean identity to isolate \( \sin x \).
• Once the equation takes the quadratic form \( 2\sin^2 x + \sin x - 1 = 0 \), solve for \( \sin x \).

Using the quadratic formula, you find the possible numerical values for \( \sin x \), which are crucial for determining the angle \( x \). Understanding quadratic equations helps to efficiently tackle and solve such transformed equations.

The sine function, often written as \( \sin(x) \), is one of the fundamental trigonometric functions. It gives the y-coordinate (or vertical position) of a point on the unit circle as it moves around that circle. The sine function is periodic, with a period of \( 2\pi \), meaning it repeats its values every \( 2\pi \) units.In our specific problem:

• We're interested in the angles where \( \sin x = \frac{1}{2} \) and \( \sin x = -1 \). Each of these corresponds to certain points on the unit circle.
• For instance, where \( \sin x = \frac{1}{2} \), the solutions are specific angles that belong to the first and second quadrants: \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).

Understanding the sine function's behavior and values is essential when determining possible solutions within a given range or interval.

Interval solutions in the context of trigonometric equations involve finding all possible values of \( x \) that satisfy the equation within a specified interval. Usually, this interval aligns with one full cycle of the trigonometric functions, such as from \( 0 \) to \( 2\pi \) for sine and cosine functions.For the given problem:

• We are tasked with finding solutions within \( 0 \leq x < 2\pi \).
• These represent all potential angles where the equation holds true in one full cycle of the unit circle.
• It's important to identify each solution’s appropriate position in this interval to understand their geometric significance on the circle.

Always be mindful of the interval when solving trigonometric equations, as it determines the number and type of solutions you are searching for.