Chapter 6: Problem 58
Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$2 \sin ^{2} u+\sin u-6=0$$
Short Answer
Expert verified
No solutions exist in the interval \([0, 2\pi)\)."
Step by step solution
01
Rewrite the Equation
The equation we want to solve is \(2 \sin^2 u + \sin u - 6 = 0\). This can be rewritten by considering a substitution where \(x = \sin u\). This transforms the equation into a quadratic form: \(2x^2 + x - 6 = 0\).
02
Solve the Quadratic Equation
To solve the quadratic equation \(2x^2 + x - 6 = 0\), we can apply the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), and \(c = -6\). Calculating the discriminant: \(b^2 - 4ac = 1^2 - 4 \times 2 \times (-6) = 1 + 48 = 49\). This gives the solutions for \(x\) as \(x = \frac{-1 \pm 7}{4}\).
03
Determine Values of X
Calculating the solutions for \(x\):- \(x_1 = \frac{-1 + 7}{4} = \frac{6}{4} = \frac{3}{2}\)- \(x_2 = \frac{-1 - 7}{4} = \frac{-8}{4} = -2\)For \(x = \frac{3}{2}\), \(x\) must be within the range \([-1,1]\) for \(\sin u\). Therefore, we only consider \(x = -2\), which is actually not within the valid range for \(\sin u\). This step indicates a need to re-evaluate, as \(\sin u = x\) is not possible with these results.
04
Verify Realistic Solutions
Since \(x = \frac{3}{2}\) and \(x = -2\) are not valid for \(\sin u\), we appear to have no solutions for this specific equation that meet the constraint \([-1, 1]\) for a sine function. Therefore, the given equation has no solutions in the interval \([0, 2\pi)\).
05
Conclusion
Having checked the validity of the calculated \(x\) values, neither lies within the interval required by \(\sin u\). Consequently, the equation \(2 \sin^2 u + \sin u - 6 = 0\) has no solutions in the interval \([0, 2\pi)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a type of polynomial equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). In solving trigonometric equations, we often transform them into a quadratic form to make use of the quadratic formula for finding solutions.
To solve a quadratic equation, you can use the quadratic formula:
To solve a quadratic equation, you can use the quadratic formula:
- \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
- If it's positive, there are two distinct real solutions.
- If it's zero, there is exactly one real solution.
- If it's negative, the solutions are complex or non-real.
Sine Function
The sine function, denoted as \(\sin\), is one of the fundamental trigonometric functions. It relates an angle in a right triangle to the ratio of the length of the opposite side to the hypotenuse. The sine function is periodic, with a period of \(2\pi\), and its range is restricted between \([-1, 1]\).
Sine function properties include:
Sine function properties include:
- Periodicity: \(\sin(x + 2\pi) = \sin x\)
- Range: \([-1, 1]\)
- Symmetrical properties: \(\sin(-x) = -\sin x\)
Interval Notation
Interval notation is a concise way of representing subsets of the real number line, which is often used in mathematics to specify the range in which solutions are considered.
For example, the interval \([0, 2\pi)\) includes all real numbers \(x\) such that \(0 \leq x < 2\pi\). The bracket \([\) indicates that the endpoint is included in the interval, while the parenthesis \()\) suggests that the endpoint is not included. This notation is particularly useful to:
For example, the interval \([0, 2\pi)\) includes all real numbers \(x\) such that \(0 \leq x < 2\pi\). The bracket \([\) indicates that the endpoint is included in the interval, while the parenthesis \()\) suggests that the endpoint is not included. This notation is particularly useful to:
- Express domains of functions, ensuring solutions fall within a valid range.
- Communicate clearly about where solutions are valid, like \([0, 2\pi)\) for periodic functions like sine.
Trigonometric Solutions
Trigonometric solutions are the values of the variable, typically an angle, that satisfy a trigonometric equation. The process of solving these involves multiple considerations:
- Transform the equation into a solvable form, which might include converting it to a quadratic equation.
- Ensure that solutions adhere to the constraints of the trigonometric functions involved. For instance, the output of a sine function must be between \([-1, 1]\).
- Verify solutions within the specified interval, often using interval notation for precision.