Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 6: Problem 67
Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$2 \sin ^{3} x+\sin ^{2} x-2 \sin x-1=0$$
Short Answer
Expert verified
The solutions are \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6} \).
Step by step solution
01
Set Substitution for Sine
Let \( y = \sin x \). Substitute \( y \) in the equation to simplify it. The equation becomes \( 2y^3 + y^2 - 2y - 1 = 0 \). We need to solve this cubic equation for \( y \), which represents \( \sin x \).
02
Find Possible Rational Roots
Use the Rational Root Theorem to test possible rational roots. The possible rational roots could be \( \pm 1 \), considering the divisors of the constant term (-1) and the leading coefficient (2).
03
Test the Possible Rational Roots
Substitute these rational roots into the cubic equation \( 2y^3 + y^2 - 2y - 1 = 0 \). By testing \( y = 1 \), we find it does not solve the equation. Testing \( y = -1 \), we find that it satisfies the equation, confirming \( y = -1 \) is a root.
04
Perform Polynomial Division
Divide the original cubic polynomial by \( y + 1 \) to factor out the root. This can be done using synthetic division, which simplifies the cubic equation to a quadratic. The result is \( 2y^2 - y - 1 \).
05
Solve the Quadratic Equation
Solve the quadratic equation \( 2y^2 - y - 1 = 0 \) using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 2 \), \( b = -1 \), and \( c = -1 \).
06
Calculate with the Quadratic Formula
Plug the values into the quadratic formula: \( y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4\times 2 \times (-1)}}{2\times 2} \). Simplifying gives \( y = \frac{1 \pm \sqrt{9}}{4} \), which results in roots \( y = 1 \) and \( y = -\frac{1}{2} \).
07
Determine Solutions for \( x \)
Recall that \( y = \sin x \). We solve for the angles where \( \sin x = -1 \), \( \sin x = 1 \), and \( \sin x = -\frac{1}{2} \) over the interval \([0, 2\pi)\). \( \sin x = -1 \) occurs at \( x = \frac{3\pi}{2} \), \( \sin x = 1 \) occurs at \( x = \frac{\pi}{2} \), and \( \sin x = -\frac{1}{2} \) at \( x = \frac{7\pi}{6} \) or \( x = \frac{11\pi}{6} \).
08
Finalize the Solution
The solutions in the interval \([0, 2\pi)\) are \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6} \).
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 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Division
Polynomial division is a method used to divide one polynomial by another. One of the most practical techniques is synthetic division, which simplifies the process, especially when the divisor is a linear factor.
- To perform synthetic division, arrange the coefficients of the polynomial in decreasing order of power.
- Identify the root of the linear factor. For example, when using the divisor \( y + 1 \), the root is \( y = -1 \).
- Run the synthetic division process by applying the root to the coefficients to obtain a new polynomial with one lower degree than the original.
Rational Root Theorem
The Rational Root Theorem is a useful tool when dealing with polynomial equations. It helps identify potential rational roots of a polynomial.
- The theorem states that any rational solution of the polynomial equation \( ax^n + bx^{n-1} + \, ... \, + k = 0 \) has the form \( \frac{p}{q} \), where \( p \) divides the constant term \( k \) and \( q \) divides the leading coefficient \( a \).
- For the problem \( 2y^3 + y^2 - 2y - 1 = 0 \), the constant term is \(-1\) and the leading coefficient is \(2\). Thus, the possible rational roots are \( \pm 1 \).
Quadratic Formula
The Quadratic Formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula provides the roots of the equation using:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Plug the coefficients \( a \), \( b \), and \( c \) into the formula. In our case, \( a = 2 \), \( b = -1 \), and \( c = -1 \).
- Substitute these values: \( y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4\times 2 \times (-1)}}{2\times 2} \).
- Simplifying gives \( y = \frac{1 \pm \sqrt{9}}{4} \). The solutions are \( y = 1 \) and \( y = -\frac{1}{2} \).
Interval Solutions
Interval solutions in trigonometric equations involve determining over which specific intervals the solutions apply. For trigonometric functions, this often entails limiting the search for solutions within a certain range, typically \([0, 2\pi)\).
- In our exercise, the roots \( y = 1 \), \( y = -\frac{1}{2} \), and \( y = -1 \) translate back to trigonometric identities for \( \sin x \).
- We then find angles \( x \) such that \( \sin x \) matches each root, within the interval \([0, 2\pi)\).
- \( \sin x = 1 \) takes place at \( x = \frac{\pi}{2} \).
- \( \sin x = -1 \) happens at \( x = \frac{3\pi}{2} \).
- \( \sin x = -\frac{1}{2} \) occurs at \( x = \frac{7\pi}{6} \) and \( x = \frac{11\pi}{6} \).
