Chapter 7: Problem 266
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 \( x = \frac{\pi}{2} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
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:
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.
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.
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.
- \( \sin 0 = 0 \)
- \( \sin \frac{\pi}{2} = 1 \)
- \( \sin \pi = 0 \)
- \( \sin \frac{3\pi}{2} = -1 \)
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:
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 \(a\).
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:
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.