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For the following exercises, solve exactly on the interval \([0,2 \pi) .\) Use the quadratic formula if the equations do not factor. $$ \sin ^{2} x+\sin x-2=0 $$

Short Answer

Expert verified
The solution is \( x = \frac{\pi}{2} \).

Step by step solution

01

Identify the Form of the Equation

The given equation is \( \sin^2 x + \sin x - 2 = 0 \). This equation is quadratic in form, where the variable is \( \sin x \). It resembles the standard quadratic equation \( ax^2 + bx + c = 0 \), with \( a = 1 \), \( b = 1 \), and \( c = -2 \).
02

Apply the Quadratic Formula

To solve \( \sin^2 x + \sin x - 2 = 0 \), we apply the quadratic formula: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \] where \( a = 1 \), \( b = 1 \), and \( c = -2 \). Substitute these values into the formula: \[ \sin x = \frac{{-1 \pm \sqrt{{1^2 - 4 \cdot 1 \cdot (-2)}}}}{2 \cdot 1} \] Simplifying further, we find:\[ \sin x = \frac{{-1 \pm \sqrt{{1 + 8}}}}{2} = \frac{{-1 \pm 3}}{2} \]
03

Solve for Possible Values of \( \sin x \)

The equation \( \sin x = \frac{{-1 + 3}}{2} = 1 \) and \( \sin x = \frac{{-1 - 3}}{2} = -2 \).- \( \sin x = 1 \) is valid, while \( \sin x = -2 \) is not valid since the sine function's range is \([-1, 1]\).
04

Find \( x \) Values in the Interval \([0, 2\pi)\)

For \( \sin x = 1 \), we need to determine when the sine function equals 1 in the interval \([0, 2\pi)\). Recall that \( \sin x = 1 \) at \( x = \frac{\pi}{2} \).Thus, the solution to the equation in the interval \([0, 2\pi)\) is \( x = \frac{\pi}{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Formula
The Quadratic Formula is a universal tool used to find the roots of quadratic equations. A quadratic equation, typically written in the form \( ax^2 + bx + c = 0 \), involves three constants: \( a \), \( b \), and \( c \), with \( a \) not equal to zero.
The formula is derived from the process of completing the square and is given as:
\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\]
With this formula, you can directly calculate the solutions of any quadratic equation. The quadratic formula is especially useful when the equation doesn't factor easily. Here, the discriminant, \( b^2 - 4ac \), plays a crucial role:
  • If the discriminant is positive, there are two distinct real solutions.
  • If it is zero, there is exactly one real solution (a repeated root).
  • If it is negative, there are no real solutions, only complex ones.
The Quadratic Formula is not only precise but also saves time on a potentially complex factorization process.
Sine Function
The sine function, denoted as \( \sin x \), is a fundamental trigonometric function crucial for understanding periodic phenomena. It measures the vertical component of a point on the unit circle as it sweeps around an origin.

This function is periodic with a period of \( 2\pi \), meaning it completes a full cycle of behavior over intervals of \( 2\pi \). Sine is an odd function, reflecting symmetrically around the origin on the graph. Its range is limited to \([-1, 1]\), which dictates that values of \( \sin x \) must always fall within these bounds.
  • \( \sin 0 = 0 \)
  • \( \sin \frac{\pi}{2} = 1 \)
  • \( \sin \pi = 0 \)
  • \( \sin \frac{3\pi}{2} = -1 \)
In solving trigonometric equations like \( \sin^2 x + \sin x - 2 = 0 \), understanding the range and period of the sine function is crucial to finding valid solutions.
Interval Notation
Interval Notation is a simple and efficient way to describe a set of numbers within a certain range. It's often used in mathematical contexts, like when solving equations, to clearly communicate solutions that fall within specified boundaries.

In Interval Notation, brackets and parentheses represent the endpoints:
  • \([a, b]\) includes both endpoints \(a\) and \(b\).
  • \((a, b)\) excludes both endpoints \(a\) and \(b\).
  • \([a, b)\) includes \(a\) but excludes \(b\).
  • \((a, b]\) includes \(b\) but excludes \(a\).
The interval \([0, 2\pi)\), frequently used in trigonometry, signifies a span that includes 0 but excludes \(2\pi\). This is particularly important if you're looking for solutions to trigonometric equations over a circular rotation of a unit circle. By using interval notation, it becomes convenient to distinguish between different parts of a solution set succinctly and precisely.
Unit Circle
The Unit Circle is a fundamental concept in trigonometry, representing a circle with a radius of 1 centered at the origin \((0,0)\) of the coordinate plane. It is crucial for understanding the behavior of trigonometric functions.

Key aspects of the Unit Circle include:
  • Angles are measured in radians. One complete revolution around the circle equals \( 2\pi \) radians.
  • The coordinates of points around the circle correspond to \((\cos \theta, \sin \theta)\), where \(\theta\) is the angle formed with the positive x-axis.
  • Special angles, like \(\frac{\pi}{2}, \pi, \frac{3\pi}{2}\), and \(2\pi\) provide reference points for computations.
The Unit Circle helps visualize why certain values of sine, cosine, and tangent repeat periodically. For example, for the angle \( \frac{\pi}{2} \), the sine value is 1, which connects back to finding solutions for trigonometric equations such as \(\sin x = 1\). Understanding the Unit Circle aids in working through and solving trigonometric equations by providing a spatial reference for angles and their corresponding function values.

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